Algebra NotesIntroduction to Algebra.........................................................................................................................4

Addition and Subtraction...................................................................................................................4

Multiplication and Division................................................................................................................5

Variables on Both Sides of The Equation.........................................................................................10

Combining Like Terms + Combining Like Terms With Distribution......................................................17

BOMDAS/BODMAS/PEMDAS..............................................................................................................23

Substitution.........................................................................................................................................25

Equations and Formulas......................................................................................................................26

Multi-Step Equations...........................................................................................................................31

Introduction to inequalities.................................................................................................................40

Solving inequalities..............................................................................................................................41

Solving inequalities: Adding or Subtracting a value.........................................................................41

Solving inequalities: Swapping the sides of the equation................................................................41

Solving inequalities: Multiplying or Dividing by a Value..................................................................42

Multiplying/dividing by a positive number..................................................................................42

Multiplying/dividing by a negative number.................................................................................42

Multiplying/dividing by Variables................................................................................................42

Solving Two Inequalities at Once.....................................................................................................42

Exponents............................................................................................................................................44

Squares and Square Roots...............................................................................................................45

Cubes and Cube Roots.....................................................................................................................46

Nth Root..........................................................................................................................................46

Multiplication and Division..........................................................................................................46

Rearranging exponents to becoming roots..................................................................................47

nth Root of a-to-the-nth-Power...................................................................................................47

nth Root of a-to-the-mth-Power..................................................................................................48

Surds................................................................................................................................................49

Fractional Exponents.......................................................................................................................51

Laws of Exponents...........................................................................................................................53

How to multiply and divide variables with exponents.....................................................................54

Expanding............................................................................................................................................56

Multiplying Negatives..........................................................................................................................57

Commutative, Associative and Distributive Laws................................................................................58

Cross Multiplication.............................................................................................................................60

Fractions in Algebra.............................................................................................................................61

Converting Repeating Decimals Into Fractions................................................................................65

Absolute Value Equations....................................................................................................................71

Absolute Values in Inequalities........................................................................................................76

Factoring in Algebra + Special Binomial Products................................................................................79

Special Binomial Products................................................................................................................79

Factoring..........................................................................................................................................81

Introduction to Logarithms..................................................................................................................85

Polynomials.........................................................................................................................................86

Adding and Subtracting Polynomials...............................................................................................88

Multiplying Polynomials..................................................................................................................91

Long Polynomial Multiplication.......................................................................................................94

Dividing polynomials........................................................................................................................98

Conjugate......................................................................................................................................111

Linear Equations................................................................................................................................119

Equation of a straight line..............................................................................................................119

Exploring the Line Graph...............................................................................................................125

Cartesian Coordinates...................................................................................................................127

Explanation....................................................................................................................................130

Rational Expressions......................................................................................................................131

Solving Polynomials.......................................................................................................................132

Quadratic Equations..........................................................................................................................133

Factoring Quadratic Equations......................................................................................................138

Completing the Square..................................................................................................................148

Functions...........................................................................................................................................153

Sequences.........................................................................................................................................163

Arithmetic Sequences (AKA Arithmetic Progressions)...................................................................164

Geometric Sequences....................................................................................................................170

Fibonacci Sequences......................................................................................................................178

Triangular Sequences....................................................................................................................178

Sequences – Finding a Rule...........................................................................................................178

Simplifying Algebraic Equations.........................................................................................................190

Combining Like Terms...................................................................................................................190

Distributive property.....................................................................................................................191

Evaluating expressions..................................................................................................................192

Writing expressions...........................................................................................................................193

Variables, Expressions and Equations................................................................................................193

Simple equations...........................................................................................................................194

Misc/Jumbled Algebra Problems and Solutions................................................................................194

Word Problems..................................................................................................................................225

Sources..............................................................................................................................................229

Introduction to Algebra

Addition and SubtractionOne of the fundamental concepts to grasp here is that whatever we do to one side of the equation, we do to the other. Saying that x + 11 = 7 is saying that they’re equal quantities. In order for this equal to hold true, anything we do to one side of the equation, we have to do to the other. To solve this, we would subtract 11 from both sides, giving us x = 4.

Examples

1. X – 2 = 4

Obviously the answer is 6, but we need to work through it step by step.

Step 1: Get rid of the “-2”. To do this, we do the exact opposite of “-2” which is “+2”. Obviously, -2 + 2 = 0.

Step 2: Add the +2 to the other side of the equation (the “4). 4 + 2 = 6.

We are now left with x + 0 = 6. Therefore, x = 6.

2. X + 5 = 12

Step 1:start with the base equation – x + 5 = 12

Step 2: subtract 5 from both sides: x + 5 – 5 = 12 -5

Step 3: do the require arithmetic: x + 0 = 7

Step 4: answer the issue: x = 7

3. 10 – y = 6

Step 1: 10 – y = 6

Step 2: 10 – 10 – y = 6 – 10

Step 3: 0 – y = -4

Step 4: - y = -4

So y = 4

4. 9 – x = 3 ½

Step 1: 9 – x = 3 ½

Step 2: 9-9-x = 3 ½ - 9

Step 3: 0 – x = 5 ½

Step 4: - x = - 5 ½

X = 5 ½

5. -9 + x = -2

Step 1: -9 + x = -2

Step 2: -9 + 9 + x= -2 + 9

Step 3 = 0 + x = 7

X = 7

Multiplication and DivisionNote: this is relatively the same process as above in that the we are still combining the opposite of the term (the number) that precedes the variable (the letter) with the “answer” (the number the solved equation will eventually equal).

Examples

1. X x 4 = 8

Step 1: In algebra, we would express “X x 4” as “4x” (looks less confusing). So we say first that “4x = 8).

Step 2: now we do what is mentioned above, that is, we do the opposite of the term to both the term and the “answer”. So we now get 4x ÷ 4x = 8 ÷ 4

Step 3: do the arithmetic = 1x (or just “x”) = 2

Therefore, x = 2.

2. X ÷ 3 = 5

Step 1: x ÷ 3 = 5

Step 2: x ÷ 3 x 3 = 5 x 3

Step 3: x ÷ 1 = 15

Step 4: 1x = 15

Therefore, x = 15

3. X ÷ 3 + 2 = 5

Step 1: x ÷ 3 + 2 = 5

Step 2: remove the “2”: x ÷ 3 + (2-2) = 5 -2

Step 3: x ÷ 3 + 0 = 3

Step 4: x ÷ 3 = 3

Step 5: now get rid of the “÷3”: x ÷ 3 x 3 = 3 x 3

Step 6: 1x = 9

Therefore, x = 9.

4. 3x – 2 = 7

Step 1: 3x -2 = 7

Step 2: 3x – (-2 + 2) = 7 + 2

Step 3: 3x – 0 = 9

Step 4: 3x ÷ 3 = 9/3

Step 5: 1x = 3

Therefore x = 3

5. 2(x + 3) = 10

Step 1: 2 (x + 3) = 10

Step 2: Divide both sides by 2: 2 (x + 3)/2 = 10/2

Step 3: Do the arithmetic: x + 3 = 5

Step 4: subtract the 3 from both sides: x + 3 – 3 = 5 – 3

Step 5: do the arithmetic: x + 0 =2

Therefore x = 2.

6. 7 + 5x = 37

Step 1: 7 + 5x = 37

Step 2: (7 – 7) + 5x = (30 – 7)

Step 3: 0 + 5x = 30

Step 4: 30 ÷ 5 = 6

Therefore x = 6

7.5x12

=23

Step 1 (write the base equation): 5x12

=23

Step 2 (isolate the ‘x’. We do this by doing the exact opposite of 5/12, which is to multiply both sides

by 12/5): x = 23x125

. You can divide the 12 by 3 and get a 4, and divide the 3 by 3 and get a 1. So we

get 21x45

= 85

Note, you can verify your answer in algebra. We got x = 8/5. Let’s check this, we said 5/12 times x

times 8/12 was equal to 2/3rds. 512

x85

Cancel the 5’s, to give us

812

then simplify by divideby 4=23

8. 1.3x – 0.7x = 12

Step 1 (write the base equation): 1.3x – 0.7x = 12

Step 2 (0.7x and 1.3x are like terms. So subtract t 0.7x from 1.3x): 0.6x = 12

Step 3 (divide 12 by 0.6): x = 20.

9. 5x – (3x + 2) = 1

Step 1 (write the base equation): 5x – (3x + 2) = 1

Step 2 (we want to work with the distributive property here, so we can add a 1 just before the brackets, which obviously becomes -1): 5x – 3x – 2 = 1.

Step 3 (subtract the 3x from 5x): 2x – 2 = 1

Step 4 (get rid of the -2 by adding it to 1): 2x = 3

Step 5 (get rid of the 2 from 2x): x = 1.5

10. x−(3 s8 )=56Step 1 (write the base equation): x−(3 s8 )=56Step 2 (factor out an ‘s’): s(1−38 )=56

Step 3 (saying 1 – 3/8 is the same as writing 8/8 – 3/8 = 5/8): 58s=56

Step 4 (isolate ‘s’ by removing the 5/8. You do this by multiplying both sides by 5/8): s = 4/3 (note

that you’d get 56x85<the 5' scancel out∧ you' releft with

86=43

11.5 (q−7 )12

=23

There are a few ways of doing this:

First method:

Step 1 (write the base equation): 5 (q−7 )12

=23

Step 2 (rearrange the equation into): 512

(q−7 )=23

Step 3 (remove the 5/12 by multiplying both sides by 12/5): (q−7 )=23x125

(divide the 12 by 3, you

get a 4, and divide the 3 by 3 you get 1. Then, 2 x 4 is 8/5).

q−7=85

Step 4 (remove the -7 by adding it to 8/5. To add fractions they need a common denominator. 5 and 7 Second method:

Step 1 (write the base equation): 5 (q−7 )12

=23

Step 2 (multiply both sides by 12): 5 (q−7 )=243

24/3 = 8. So we get5 (q−7 )=8

Step 3 (distribute the 5): 5q−35=8

Step 4 (get rid of -35): 5q = 43

Step 5 (get rid of the 5 by multiplying both sides of the equation by 1/5): q = 431x15=43/5

Khan’s method:

12. Jade is stranded downtown with only $10 to get home. Taxis cost $0.75 per mile, but there is an additional $2.35 hire charge. Write a formula and use it to calculate how many miles she can travel with her money. Determine how many miles she can ride.

Cost equals the initial hire cost + 0.75 x the number of miles. The cost is $10, so cost has to be $10.

1. 10 = 2.35 + 0.75m.2. Subtract 2.35: 7.65 = 0.75m3. Divide by 0.75: 10.24. So she can travel 10.2 miles, enough to get her home.

Word problemThe perimeter of Tina’s rectangular garden is 60 feet. If the length of the garden is twice the width, what are the dimensions of the garden?

The formula for a perimeter is w + w + l + l.

Length will = 2w because the width is twice the width. So perimeter will equal w + 2w + w + 2w = 6w.

So 6w = 60 feet. Divide 60 by 6 = 10. Therefore w = 10. Therefore, 2w = 20.

Variables on Both Sides of The EquationTake the example problem of 2x + 3 = 5x – 2. We are still trying to isolate x. Start off by subtracting 2x from both sides of the equation. This gives us 3 = 3x – 2. So we want to get rid of the -2 now. Add it to both sides of the equation. We get 5 = 3x. Remove the 3, which gives us 5/3 = x. We could also write this as a mixed number. 3 goes into 5 once, with a remainder of 2, so we can write 1 2/3 = x

Examples

1. 20 – 7y = 6y – 6

Step 1 (write the base equation): 20 – 7y = 6y – 6

Step 2 (remove the 7y by adding it to both sides of the equation): 20 = 13y – 6

Step 3 (remove the -6 by adding it to the left hand side): 26 = 13y

Step 4 (isolate y by removing 13. Do this by dividing 26 by 13): 2 = y OR y = 2

NOTE you could do this another way too.

Step 1 (write the base equation): 20 – 7y = 6y – 6

Step 2 (remove the 6y by subtracting it on both sides): 20 – 13y = -6

Step 3 (subtract 20 from both sides): -13y = -26

Step 4 (isolate y by dividing both sides by -13): y = 2

NOTE you can verify your answer by putting y back into the equation

Step 1 (substitute y for 2): 20 – 7 x 2 = 6 x 2 – 6

Step 2 (do the arithmetic): 20 – 14 = 12 – 6 = 6 = 6

2. Solve for x: 8 (3x + 10) = 28x – 14 – 4x

Step 1 (write the base equation): 8 (3x + 10) = 28x – 14 – 4x

Step 2 (distribute the 8 into the brackets): 24x + 80 = 28x – 14 – 4x

Step 3 (combine the 28x and -4x = 24x): 24x + 80 = 24x – 14

Step 4 (isolate 24x by subtracting on both sides): 80 = -14 THEREFORE this equation has no solution, 80 =\= -14

3. 6 x − 2 = 2 x + 3

Step 1 (write the base equation): 6x – 2 = 2x + 3

Step 2 (remove the 2x): 4x – 2 = 3

Step 3 (remove the -2): 4x = 5

Step 4 (divide the 5 by 4): x = 5/4

4. − 5 qr − r + 7 s − 5 = 9 r − 3 s + 4 Solve for q

Combining Like Terms + Combining Like Terms With DistributionNote: the variable should be on the left.

BOMDAS/BODMAS/PEMDASBOMDAS means “Brackets, Orders, Multiplication, Division, Addition and Subtraction”. This indicates the order in which issues should be “figured out”.

From “Math is Fun” website:

Brackets firstOrders (ie Powers and Square Roots, etc.)Division and Multiplication (left-to-right)Addition and Subtraction (left-to-right)

Examples

1. 3 + 6 x 2

Step 1: We do the multiplication section first, which is 6 x 2 = 12. So we now have 3 + 12

Step 2: Then we do the addition, which finishes the equation. 3 + 12 = 15

2. (3 + 6) x 2

Step 1: Brackets come first, so 3 + 6 = 9. We now have 9 x 2

Step 2: Then do the multiplication. 9 x 2 = 18.

3. 12 ÷ 6 x 3 ÷ 2

Remember that in terms of “order” or “priority”, multiplication and division rank equally. In this case, we only have multiplication and division, so we simply work from left to right.

Step 1: 12 ÷ 6 = 2. So we now have 2 x 3 ÷ 2.

Step 2: 2 x 3 = 6. So we now have 6 ÷ 2.

Step 3: 6 ÷ 2 = 3.

4. 3 + 6 ÷ 3 x 2

Step 1: 6 ÷ 3 = 2. So we now have 3 + 2 x 2

Step 2: 2 x 2 = 4. So we now have 3 + 4.

Step 3: 3 + 4 = 7.

5. 30 – (5 x 2 3 – 15)

Brackets come first, but when working within brackets, remember the B OMDAS still applies.

Step 1: 23 = 8. So we now have 30 – (5 x 8 – 15)

Step 2: 5 x 8 = 40. So we now have 30 – (40 – 15)

Step 3: 40 – 15 = 25. So we now have 30 – 25.

Step 4: 30 – 25 = 5.

6. (4 2 - 6 + 5) ÷ (3 2 + 8 - 7 × 2)

Step 1: First bracket: 42 – 6 + 5 = 16 – 6 + 5.

Step 2: 16 – 6 + 5 = 10 + 5.

Step 3: 10 + 5 = 15.

Step 4: start second bracket: 32 + 8 – 7 x 2 = 9 + 8 – 7 x 2.

Step 5: 9 + 8 – 7 x 2 = 9 + 8 – 14.

Step 6: 9 + 8 – 14 = 17 – 14.

Step 7: 17 – 14 = 3.

Step 8: divide bracket 1 by bracket 2: 15/3 = 5.

Substitution“Substitute” means to put in the place of another. For the purposes of algebra, it basically means putting numbers where the letters are.

Remember that:

Two like signs become a positive signo Adding two positive numbers will result in a positive outcome. E.g. 3 + (+2) = +5.o Subtracting a negative number from a positive will become a positive. E.g. 6 – (-3) =

6 + 3 = 9o Multiplying a positive by a positive becomes a positive. E.g. 3 x 2 = 6.o Multiplying a negative and a negative becomes a positive. E.g. (-3) x (-2) = 6o The square-root of a negative number will be positive. E.g. -42 = +16

Two unlike signs become a negative sign:o Adding a positive number to a negative number results in a negative sign. E.g. 7 + (-

2) = 7 – 2 = 5.o Subtracting a negative number from a positive number results in a negative sign. E.g.

8 – (+2) = 8 – 2 = 6.o Multiplying a positive number by a negative one results in a negative number. E.g. 3

x (-2) = -6.

Examples

1. 10/x + 4 where x = 5.

Step 1: put the “5” where the “x” is: 10/5 + 4

Step 2: 2 + 4 = 6.

2.

Step 1: 5(10-4)/2

Step 2: 5(6)/2

Step 3: 30/2 = 15.

3. xy + y 2 where x = 2 AND y = 3

Step 1:2x3 + 32

Step 2: 6 + 9 = 15

4. z 2 – z – 1 where z = -3

Step 1: (-3)2 (-3) -1

Step 2: 9 + 3 -1

Step 3: 12 -1 = 11.

5. 3y 2 + 7y – 4 where y = - 5

Step 1: 3 x (-5)2 + 7 x -5 – 4

Step 2: 75 + 7 x -5 – 4

Step 3: 75 - 35 – 4

Step 4. 40 – 4 = 36

Equations and FormulasAn equation is something that says two things are equal, therefore an equals sign “=” is always present. An example would be x + 2 = 6. We are saying that the left hand side of the equation (x + 2) is equal to the right hand side of the equation (6)

A formula on the other hand is a type of equation that represents the relationship between different variables (which are letters, like x and y – these represent numbers that we don’t know yet). Formulas have more than one variable. An example of a formula is as follows:

V = hwl

V stands for volume, h for height, w for width, and l for length.

4 x 5 x 10 = 200

But note that a formula can be written without the “=”, so we could just write ‘hwl’ for volume of the box.

Generally, the ‘subject’ of a formula is a single variable that everything else is equal to (see above).

Changing the subject (rearranging)

In algebra, we can “rearrange” a formula so that another variable becomes the ‘subject’. For example, we know that (V = hwl) is the formula for finding the volume of a box. Currently “V” is the subject, but let’s try to make “w” the subject.

To rearrange, we do the following steps:

Step 1 (Start with the base formula): v = hwl

Step 2 (Divide both sides by “h”; this means we divide both left hand side (v) and right hand side (hwl) by h): V/h = wl

Step 3 (Divide both sides by “l”): V/hl = w

Step 4 (Swap sides): w = V/hl

So, in summary, to rearrange that type of formula, divide the original subject by all of the variables except the variable we wanted to make the new subject.

Examples

1. For the formula x = 3y – z, what is the value of x when y=4 and z =1?

Step 1 (write the base formula): x = 3y – z

Step 2 (substitute the variables for the known numbers): x = 3 x 4 – 1

Step 3 (working in order, remembering BOMDAS): x = 12 -1

Step 4 (solve): x = 11

2. For the formula v 2 = u 2 + 2ad, what is the value of v when u = 15, a = 10 and d = 20

Step 1 (write the base formula): v2 = u2 + 2ad

Step 2 (substitute the variable for the known numbers): v2 = 152 + 2x 10 x 20

Step 3 (work the right hand side of the equation out in the correct order, remembering BOMDAS): 225 + 20 x 20

Step 4 (keep working through, you could have done all of this in step 3, but this is just a reminder for the BOMDAS steps): 225 + 400 = 625.

3. The formula for the total surface area of a cylinder of base radius “r” and height “h” is: A = 2πr 2 + 2πrh. What is the value of “A” when π = 3.14, r = 7 and h = 10?

Step 1 (write the base formula): 2πr2 + 2πrh

Step 2 (Substitute the variables for the known numbers) 2 x 3.14 x 72 + 2 x 3.14 x 7 x 10.

Step 3 (work through in order, remembering BOMDAS): 307.87 + 439.82 = 747.69 (round up to 748)

4. Rearrange this formula, to make “b” the subject: P = 2a + 2b

Step 1 (write the base formula): P = 2a + 2b

Step 2 (as we are making “b” the subject, we move “2a” to the other side. To do this we subtract it from both sides). P – 2a = 2a + 2b – 2a

Step 3 (simplify the above equation): P – 2a = 2b

Step 4 (swap sides): 2b = p – 2a

Step 5 (now divide both sides by 2): 2b/2 = (P – 2a)/2

Step 6: b = (P -2a)/2

5. Rearrange y = bx – c so that x becomes the subject.

Step 1 (write the base formula): y = bx – c

Step 2 (as we want to make x the subject, we want to isolate the term that has “x” in it, which is “bx”, so we need to get rid of the “-c”. So, we add “c” to both sides, which gives us): y + c = bx

Step 3 (we still haven’t isolated “x” yet, so we use the ‘multiplication principle’, and as it is “bx”, we

divide both sides of the equation by b): y+cb

=x

6. Rearrange the following formula for the area of a triangle so that “b” becomes the

subject: A=12bh

Step 1 (write the base formula): A=12bh

Step 2 (as we want to make “b” the subject, we need to get rid of everything that isn’t “b”. We can

start with the fraction first. To get rid of 12

we multiply both sides of the equation by 2, and as 2 x

12=1 we don’t need to write the 1, therefore we have removed the

12

which gives us): 2A = bh

Step 3 (we still need to isolate “b”, so we have to get rid of “h” by dividing both sides by “h”): 2 Ah

=b

7. Rearrange the following formula so that “b” becomes the subject: A = 2a 2 + 4ab.

Step 1 (write the base formula): A = 2a2 + 4ab

Step 2 (get rid of the first term that has nothing to do with “b”, which is 2a 2. As “2a2” is a positive number, we do the exact opposite which is subtraction. Therefore, we subtract “2a2” from both sides): A – 2a2 = 4ab

Step 3 (we still need to isolate “b”, which is part of “4ab”. To do this, we get rid of “4a”. As “4a” is being multiplied into “b”, we do the exact opposite, which is division. Therefore, we divide “4a”

from both sides: A−2a2

4a=b

8. Rearrange the following formula so that “c” becomes the subject: A= c2

4 π

Step 1 (write the base formula): A= c2

4 π

Step 2 (get rid of the term that has nothing to do with “c”, which is 4π. As “4π” is dividing into c2, we do the exact opposite which is multiplication): 4πA = c2

Step 3 (isolate “c” by getting rid of the power. As the power is squaring “c”, we do the exact opposite by deriving the square root on both sides): √4 πA = c

9. Rearrange the following formula that is used in physics to calculate distance, so that “a”

becomes the subject: s = ut + 12

at 2

Step 1 (write the base formula): s = ut + 12

at2

Step 2 (we don’t go straight for the fraction as we did in the previous example, we are working from left to right here, so, get rid of the positive “ut” by doing the exact which involves subtracting): s – ut

= 12

at2

Step 3 (get rid of the faction now. We do this by doing the complete reverse of the fraction. As

multiplying anything by 12

has the effect of halving it, we do the exact opposite by reversing the

fraction to 21

to double both sides ): 2 (s – ut) = at2

Step 4 (we want to isolate “a” from the term “at2”, so we need to get rid of “t2”. As “t2” is being

multiplying “a”, we do the exact opposite and divide it on both sides): 2 ( s−ut )

t 2=a

10. Rearrange the following formula so that “b” becomes the subject: P = 2a + 2b

Step 1 (write the base formula): P = 2a + 2b

Step 2 (get rid of “2a”): P – 2a = 2b

Step 3 (get rid of the “2” from “2b”): P−2a2

=b

11. y = 2x – 7. Make “x” the subject

The answer is x = (y + 7)/2. This means that we should remove the “7” first.

12. W = 23v so that v is the subject

Step 1 (write the base formula): W = 23v

Step 2 (get rid of the fraction by flipping it upside down HOWEVER, note that this is slightly different

to the times we’ve seen example when the fraction was 12

. Now, we multiply the current subject

(which is “w”) by the denominator of the fraction (which is “3”) and divide it by the numerator

(which was “2”): v=3w2

13. The formula for converting a Celsius temperature into Fahrenheit is C=59

(F−32 )

Rewrite the formula so that it solves for Fahrenheit

Step 1 (write the base formula): C=59

(F−32 )

Step 2 (remove the 5/9 by multiplying both sides by 9/5): 95C=F−32

Step 3 (remove the – 32 by adding 32 to both sides): F=95C+32

14.

15.

16.

Multi-Step EquationsYou have the necessary skills to do this, it just may not seem quite so intuitive at first. This section will be purely examples:

1.14 x+4−3x−2

=8.Solve for x

Step 1 (write the base equation): 14 x+4−3x−2

=8

Step 2 (multiply both sides by (-3x – 2): 14x + 4 = -24x – 16

Step 3 (add 24x to both sides): 38x + 4 = -16

Step 4 (subtract 4 from both sides): 38x = - 20

Step 5 (divide both sides by 38): x = -20/38 = -10/19

2. 7−10x

=2+ 15x

Step 1 (write the base equation): 7−10x

=2+ 15x

Step 2 (multiply both sides by x): 7x – 10 = 2x + 15

Step 3 (add 10 to both sides) 7x = 2x + 25

Step 4 (subtract 2x from both sides): 5x = 25

Step 5 (divide both sides by 5): x = 5

3.34x+2=3

8x−4

Step 1 (write the base equation): 34x+2=3

8x−4

Step 2 (multiply both sides by a number to get rid of the fractions. The smallest number you could use to get whole numbers instead of fractions is 8. This is because the least common multiple of 4 and 8 is eight. So multiply both sides by 8. Remember that when you multiply one side of the equation b y a number, you have to do it to all numbers on that side): 6x + 16 = 3x – 32

Step 3 (subtract 3x from both sides): 3x + 16 = -32

Step 4 (subtract 16 from both sides): 3x = -48

Step 5 (divide both sides by 3): x = -16

4.

Introduction to inequalitiesAn inequality lets us know about the relative size of two values. You don’t have to know the exact number; just that one is greater than the other. For example, Car A and B are in a race. Car B wins. We don’t know how fast car B went, we just know it won, so we can say that B > A. We call these things inequalities because they are not equal.

Symbol What it means Example> Greater than 5 > 2< Less than 7 < 9≥ Greater than or equal to X ≥ 1≤ Less than or equal to Y ≤ 3

Solving inequalitiesWhen we talk about “solving” an inequality, our aim is to have the variable (e.g. x) on its own on the left of the inequality sign. So instead of saying that 5 > x, we could say x < 5.

We solve an inequality just as we would with our normal algebraic equations, but sometimes, we have to change the direction of the inequality so that “<” might become “>”.

Here’s what you can do without changing the direction of the inequality (see more information below)

Add/subtract a number from both sides Multiply/divide both sides by a positive number Simplify a side of the equation.

o E.g. “3x < 7 + 3” can be simplified to “3x < 10”

Doing the following will change the direction of the inequality (see more information below)

Multiply/divide both sides by a negative number Swapping left and right hand sides

o E.g. ≤ “2y + 7 < 12” could be swapped to “12 > 2y + 7”

Solving inequalities: Adding or Subtracting a valueJust like we learned above, we can solve an inequality by adding/subtracting numbers from both sides of the equation.

For example, if we were to solve: x + 3 < 7, and we subtracted 3 from both sides we would get “x < 4”.

This works for both adding and subtracting as both sides of the equation are rising/falling by the exact same amount, so they keep their positions of greater/less than relative to each other.

Alex has more coins than Billy. If both Alex and Billy get three more coins each, Alex will still have more coins than Billy.

Solving inequalities: Swapping the sides of the equationIf the variable is on the right after we’ve done the rest of the solving of the equation, we need to change sides, which means reversing the sides of the equation.

For example: 12 < x + 5. If we subtract “5” from both sides we get “7 < x” (note: “sides” refers to sides of the inequality. HOWEVER we’re not finished yet, we need the variable on the right hand side, so we simply flip the equation so that we get “x > 7”. Originally we were saying that “7 is less

than x”, now we are saying that “x is greater than 7”; these are just two different ways of saying the exact same thing.

Solving inequalities: Multiplying or Dividing by a ValueJust like we do with multiplication (above) we can multiply/divide both sides of the inequality by either a positive or negative number.

Multiplying/dividing by a positive numberSolve 3y < 15. If we divide both sides by 3 we get y < 5, which becomes our solution

Multiplying/dividing by a negative numberWhen multiplying/dividing by a negative number we have to reverse the inequality.

Example: -2y < -8. Divide by -2.

First, we start off with the base equation: -2y < -8

Second, do the division on both sides: -2y/-2 > -8/-2

Third: y > 4

Multiplying/dividing by VariablesWe can multiply/divide by a variable IF we know whether or not that variable is positive or negative.

Example: bx < 3b. Divide by “b”, assuming it is positive 1 . We would get: x < 3

Example: bx < 3b. Divide by “b”, assuming it is negative 1. This would give us “–x < -3”, which would become “x > 3” if we were to get rid of the negatives.

Solving Two Inequalities at OnceExample: -2 < (6-2x)/3 < 4

Step 1 (write the base equation): -2 < (6-2x)/3 < 4

Step 2 (clear out the “/3” by multiplying every side by 3): -6 < 6-2x < 12

Step 3 (clear out the positive 6 by subtracting 6 from each part): -12 < -2x < 6

Step 4 (clear out the “-2” by multiplying each part by -0.5 note because we are multiplying by a negative number, inequalities change their direction): 6 > x > -3. That is the solution, but we can make it neater by making smaller numbers on the left and larger ones on the right, which would look like this: “-3 < x < 6”.

Examples

1. 5x > 10

Step 1 (write the base equation): 5x > 10

Step 2 (as the 5 is multiplying the “x” we do the opposite and divide each side by 5): x > 2. That is our solution

2. 3x + 2 > 8

Step 1 (write the base equation): 3x + 2 > 8

Step 2 (get rid of the “+2” by subtracting from both sides): 3x > 6

Step 3 (get rid of the “3” by dividing both sides by 3): x > 2

3. 3x – 7 < 5

Step 1 (write the base equation): 3x – 7 < 5

Step 2 (get rid of the “-7” by adding it to both sides): 3x < 12

Step 3 (get rid of the “3” by dividing it into both sides): x < 4

4. 3(4 – y) ≥ 9

Step 1 (write the base equation): 3(4 – y) ≥ 9

Step 2 (get rid of the “3” first, by dividing it into both sides): 4 – y ≥ 3

Step 3 (get rid of the “4” by subtracting it from both sides): -y ≥ -1 (note: we would get “-y as subtracting 4 from both sides would give us “4 – y – 4”.

Step 4 (make the sides positive numbers by dividing each side by “-1”, which will reverse the inequality sign): y ≤ 1

5. -4x > -12

Step 1 (write the base equation): -4x > -12

Step 2 (divide each side by “-4”, which will reverse the inequality sign): x < 3

6. -5y – 7 ≤ 3

Step 1 (write the base equation): -5y – 7 ≤ 3

Step 2 (remove the 7 by adding it onto both sides): -5y ≤ 10

Step 3 (remove the “-5” by dividing it into both sides, which will reverse the inequality sign): y ≥ -2

7. – 4 ≤ 3x + 2 < 5

Step 1 (write the base equation): – 4 ≤ 3x + 2 < 5

Step 2 (remove the “+2” by subtracting it from each side): -6 ≤ 3x < 3

Step 3 (remove the “3” from the “3x” by dividing it into each side): -2 ≤ x < 1

8. 3 ≤ -6 - 5x < 12

Step 1 (write the base equation): 3 ≤ -6 - 5x < 12

Step 2 (remove the “-6” by adding positive 6 onto both sides): 9 ≤ -5x < 18

Step 3 (remove the “-5” by dividing it into both sides, which will reverse the inequality signs): -1.8 ≥ x > -3.6

ExponentsExponent = Index (indices)/powers

E.g. 53 = 5 x 5 x 5 = 125 OR 5^3

Negative exponent = Dividing. It tells us how many times we divide 1 by the other number.

E.g. 5-3 = 1 ÷ 5 ÷ 5 ÷ 5 = 0.008 OR 1 ÷ (5 x 5 x 5) = 0.008 OR 1/53 = 0.008

If we are given a negative exponent, like 10-3, it is easier to find the reciprocal of the positive exponent, which means 1 (one) ÷ positive exponent, which in this case would be 1/103

Reciprocal = (number)/1.

If the exponent is 1, or 0

If the exponent is 1, then all you end up with is the number itself e.g. 91 = 9

If the exponent is 0, then you get 1 e.g. 90 = 1

If you have 00, it could be either 1 or 0, so we say that it is “indeterminate”

Grouping

We can group with parentheses “()”. The following table is an example of how to solve grouping issues.

Examples

1. What is the value of (-2) -5 ?

Step 1 (write the base equation): (-2)-5

Step 2 (find the reciprocal): 1

(−2 )5 = -0.03125

2.

Step 1 (write the base equation): 1/x2

Step 2 (turn the equation upside down): x2 /1

Step 3 (simplifying): x2

3. If the reciprocal of 3y - 7 is ½, what is the value of y?

Step 1 (write the base equation): 3y – 7

Step 2 (because we’ve been told what the reciprocal of the equation is, we need to find the reciprocal of the reciprocal to find out what the equation currently equals): reciprocal of ½ = 2. Therefore 3y – 7 = 2. We can now do our basic rearranging from here.

Step 3: 3y = 9.

Step 4: y = 3.

4. If y is the reciprocal of 2, what is the difference between the reciprocal of 3y and the reciprocal of 5y?

Step 1 (write the base equation)

Step 2 (figure out what y is equal to. Here we know the reciprocal of y, so we just do the opposite, which is ½)

Step 3 (now that we know what y is equal to, we can figure out the reciprocal of 3y. 3y is currently equal to 3/2 which is the same as 3 x 0.5, or half of 3. We do the exact opposite which is 2/3).

Step 4 (find the reciprocal of 5y, repeating step 3): 5y is currently 5/2, so becomes 2/5.

Step 5 (find the difference between the two reciprocals): 23−25=1015

− 615

= 415

(remember we need

the denominators to be the same number before we can do any subtraction, so we find the lowest common denominator by finding the first number that both denominators divide equally into, and multiplying the numerators by this number. Here, they both divided into 15 first. 3 goes into 15 5 times, so that’s how we got the 10. 5 goes into 5 3 times, so that’s how we got the 6).

Squares and Square RootsSquare = a number multiplied by itself

Square root = the opposite of the square

A negative that is squared becomes a positive. E.g. -52 = (-5) x (-5) = 25 (remember that a negative times a negative = positive).

Radical = square root symbol (√)

Perfect squares = squares of whole numbers (e.g. 3, 4, 5 etc.)

Radicand = the thing you are finding the root of

Sometimes you may get very long numbers that are irrational e.g. 3.170723… An irrational number is one that is not rational, which means it cannot be expressed as a fraction. 0.333 is a rational number because it can be written as 1/3.

Examples

1. Between which pair of whole numbers does the square root of 20 lie?

Step 1: √20 = 4.47. Therefore, it lies between whole numbers “4” and “5”

Cubes and Cube RootsCube = a number multiplied by itself 3 times. E.g. 23 = 8.

Cube root = opposite of a cube. A “degree” of 3 will be present on the radical symbol. E.g.

Perfect cubes = cubes of whole numbers

Nth Root“nth” refers to a variable number used as the degree in a square root problem.

It is a general way of talking about roots, therefore it could be the 2nd, 9th or 324th and so on.

Sometimes you will be faced with problems like this:

What is “n” in this equation? . 54 =625

Multiplication and Division

You can “pull apart” terms being multiplied under the root sign like this:

Note: that “dot” is called an interpunct/middledot.1 It means “the product of” or, “multiplication”

Multiplication

E.g. (the “4” in the end simplification comes from the fact that 43 = 64)

Division

Division works in a fairly similar way to multiplication above:

Note: (a≥0 and b>0): (b cannot be zero or you would be dividing by zero)

e.g.:

Addition and subtraction

Note that the same rules for multiplication and division DO NOT work for addition/subtraction.

THESE DO NOT WORK, THEY ARE NOT VALID EXAMPLES

E.g. The Pythagoras Theorem = a2 + b2 = c2. Rearrange so that “c” is the subject.

Step 1 (write the base equation): a2 + b2 = c2

Step 2 (remove the exponent from “c” by doing the exact opposite, which is making it a radical/root): √(a2 + b2) = c

Rearranging exponents to becoming roots

If then (when n is even b must be ≥ 0 why? Because if n is positive, and b is also positive, as there is no “-“ sign next to it, this is effectively giving us two positives, which = a positive).

E.g.

1 http://en.wikipedia.org/wiki/Interpunct#In_mathematics_and_science

nth Root of a-to-the-nth-PowerWhen a value (under the radical sign) has an exponent of n, and you are also taking the nth root,

then you will get the value back again provided a ≥ 0

e.g.

Examples

Where the exponent is positive or 0 (this is exactly the same as directly above):

Where the exponent is odd: this works in exactly the same way as when the exponent is positive or 0

e.g.

However where the value is negative but the exponent is positive, we get a different formula:

where |a| = the absolute value. That is, any negatives become positive.

E.g. Note: here, there is no “degree” on the radical, as the default “degree” is “2”, so it is redundant to note that it is being squared.

Another example:

Quick reference table:2

nth Root of a-to-the-mth-PowerThis refers to the situation where the exponent and root are different values (m and n). In this

situation, this is the base formula: Note: each side gives you the EXACT same

result. E.g.

HOWEVER there is another method of doing this, and that is creating a fractional exponent. We get

the following formula: . We get this formula because the nth root is the same as an

exponent of (1/n). E.g. . Another way of thinking about this is

2 http://www.mathsisfun.com/numbers/nth-root.html

remembering that when we rearrange, we do the opposite (as we talked about above). We want to rearrange the root, so the opposite of that would be to make it an exponent. However, we already have an exponent, so we divide that root into that exponent to make a fractional exponent.

Another example: 2½ = √2

Examples

1. If the nth root of 1,296 is 6, then what is the value of n?

For this question, we just do trial and error, you would find that 64 = 1,296

2. What is "n" in this equation?

Just do trial and error, as above, and you’d find that 75 is the answer.

3.

Step 1 (Write the base equation):

Step 2 (split the numerator and denominator into their own cube roots):

Step 3 (square this fraction): ( 56 )2

= 2536

SurdsSurd = a number that can’t be simplified to remove a square root (or cube root etc.). It is essentially an irrational root. For example, the square root of 4 is NOT A SURD because it can be simplified further. However the square root of 2 is a surd because is cannot be simplified further.

Examples

1. Which of the following is not a surd?

√(35) = 5.91607...√(36) = 6 exactly√(37) = 6.08276...√(38) = 6.16441...

So √(36) is not a surd

2. How many of these numbers are surds

3. How many of these numbers are surds?

4. For which of the four right-angled triangles is the length of the diagonal not a surd?

The length of the hypotenuse (c) of a right-angled triangle is found by using the Theorem of Pythagoras: c2 = a2 + b2

In A, c2 = 32 + 32 = 9 + 9 = 18 ⇒ c = √18 = 4.24..., which is a surd.

In B, c2 = 42 + 32 = 16 + 9 = 25 ⇒ c = √25 = 5, which is not a surd.

In C, c2 = 42 + 42 = 16 + 16 = 32 ⇒ c = √32 = 5.65..., which is a surd.

In D, c2 = 52 + 42 = 25 + 16 = 41 ⇒ c = √41 = 6.40..., which is a surd.

Fractional ExponentsWhole number exponents are powers that are whole numbers. E.g. 2, 3, 4, 5

etc.

Fractional exponents are exponents that are not whole numbers, but are expressed as fractions, eg. ½. A fractional exponent can be rearranged into the radical. E.g.

Why does it become the radical? Take the example of 91/2 x 91/2. To work through this we take the following steps: 9½ × 9½ = 9(½+½) = 9(1) = 9

Another example: x1/4 = x¼ × x¼ × x¼ × x¼ = x(¼+¼+¼+¼) = x(1) = x

Complicated fractions

So, we may encounter “complicated” fractions like 43/2. What do we do there? Well, a fraction like m/n could be broken down into m x (1/n). Here’s an example. Say we had 5/6. That would equal

0.833333… We would get the same answer if we did 6 x ( 15 )Keep this in mind because it can help us turn a “complicated” fraction into the radical. Because of

mn=mx ( 1n ) we can do this:

Why? Here’s the breakdown. We already saw that mn=mx ( 1n ). How did we get to the third stage “

(xm )1n”? Because it gives us the exact same answer as the step before. For example, both the

following equations give an answer of 10.33: 76 x 15 AND (76 )

15

So then we finally simply this all into n√m

Keep in mind that the order of the fraction doesn’t matter, so it will also work for m/n = (1/n) × m:

Examples

1. What is the value of 25 1/2 ?

Step 1 (write the base equation): 2512

Step 2 (re-arrange): √25 = 5 (note that a half power is the same as a square root)

2.

Step 1 (write the base equation): (1,024 )25

Step 2 (re-arrange): 5√1,0242 = 16

Note the second way of doing this:

3.

Step 1 (write the base equation): √( 8164 )3

Step 2 (Change the order of the exponents around using)

Step 3 (work out the square root):

Step 4 (cube the answer from step 3):

Laws of ExponentsLaw Example Explanation

x1 = x 61 = 6x0 = 1 70 = 1

x-1 = 1/x 4-1 = 1/4xmxn = xm+n x2x3 = x2+3 = x5 x2x3 = (xx)(xxx) = xxxxx = x5.

xm/xn = xm-n x6/x2 = x6-2 = x4 x4/x2 = (xxxx) / (xx) = xx = x2

Note: remember that x/x = 1, so every time you see an x above and below the dividing line, you can cancel them out.

( xy )−m

= ( yx )m

an x a−n=1 52 x 5-2 = 1 a−n= 1

an Therefore an x a−n =

an x1

an = 1

(xm)n = xmn (x2)3 = x2×3 = x6 First you are multiplying x ‘m’ number of times. You then have to multiply that number

n more times.

(x3)4 = (xxx)4 = (xxx)(xxx)(xxx)(xxx) = xxxxxxxxxxxx = x12

(xy)n = xnyn (xy)3 = x3y3 Example: (xy)3 = (xy)(xy)(xy) = xyxyxy = xxxyyy = (xxx)(yyy) = x3y3

(x/y)n = xn/yn (x/y)2 = x2 / y2 Example: (x/y)3 = (x/y)(x/y)(x/y) = (xxx)/(yyy) = x3/y3

x-n = 1/xn x-3 = 1/x3

And the law about fractional exponents

Examples

1. What is (x ½ ) 6 equal to?

Step 1 (Write the base equation): (x½)6

Step 2 (keep in mind that (xm)n = xmn. X1/2 x 6 = x3

2. What is (2 y 23 )3

equal to?

Step 1 (Write the base equation): (2 y 23 )3

Keep in mind that “(y2/3)” is its own term, that’s why we can separate it from the “2” to use the law of (xm)n = xmn

How to multiply and divide variables with exponentsA variable with an exponent is something like ‘y2’.

Multiplying

When multiplying variables with exponents, you will often just add the exponents together. A variable with no exponent is still added as “1”. For example:

(x3y5)(x2yz) = x3+2y5+1z1 = x5y6z

Note that, in the above example, there were no constants (numbers like 3, 2, 7, 102, 0.5 etc.) If you have constants, simply multiply the constants while still adding the exponents. Example:

(2xy)(4y) = 2 x 4 x xy x y = 8xy2

Division

With division, simple subtract the exponent on the numerator from the exponent on the

denominator. For example: y5

y2= y5−2= y3

Here’s a bigger example: x3 y z2

x y2 y2=x3−1 y1−2 z2−2=x2 y−1 z0= x2

y

Examples

1. Simplify (2x 5 y 2 )(3y 3 )

Step 1 (write the base equation): (2x5y2)(3y3)

Step 2 (find the constants and like variables to multiply/add together): (2 × 3) × x5 × (y2 × y3) = 6 × x5 × y5 = 6x5y5

2. Simplify (30x 6 y 4 z 2 ) ÷ (5x 2 y 2 z 5 )

Step 1 (write the base equation): (30x6y4z2) ÷ (5x2y2z5)

Step 2 (find the constants and like variables to multiply/subtract): (30 ÷ 5), these are our constants. This gives us 6. Next, we have the variable ‘x’, which gives us (x6 ÷ x2), and keeping in mind we subtract the exponents here as this is a division issue, we end up with x4. The next variable is ‘y’, so we have (y4 ÷ y2), subtracting the exponents like we did previously, we get y2. Last, we have ‘z’, which gives us (z2 ÷ z5) which equals z-3.

Step 3 (put it all together): 6x4y2z-3

3. Simplify (10x 3 y) × (2xy 6 ) ÷ (4x 2 y 3 )

Step 1 (write the base equation): (10x3y) × (2xy6) ÷ (4x2y3)

Step 2 (remembering BOMDAS/PEMDAS, we do the multiplication terms first, so we are working through ‘(10x3y) × (2xy6)’ before we do the division): so we get 20x4y7.

Step 3 (do the division): (20\4) x x4-2 x y7-3 = 5x2y4

4. Simplify

Step 1 (write the base equation):

Step 2 (simplify the denominator first):

Step 3 (divide the numerator by the denominator. Remember that, because this is a division issue

now, we subtract the exponents):

Expanding‘expanding’ refers to removing the brackets ‘( )’, but not necessarily right away.

Whatever is inside the brackets needs to be multiplied/divided by what is outside the parentheses.

Example: Expand 3 x (5+2). As this is a multiplication issue for the brackets, you multiply both the 5 AND the 2 by the 3. So we get 3 x 5 + 3 x 2 = 15 + 6 = 21.

Very simple formula is: a(b+c) = ab + ac

yx (x – a) = yx2 – yxa

a (2 – b + c) = 2a –ab + ac

Examples

1. Expand 3 (x + 6)

Step 1 (write the base equation): 3(x + 6)

Step 2 (apply the ‘3’ outside the brackets to both terms): 3x + 18

2. Expand -3(a – 5)

Step 1 (write the base equation): -3(a – 5)

Step 2 (apply the ‘-3’ to the terms in the brackets): -3a + 15. Note that we get ‘+15’ as we are multiplying a negative 3 by a negative 5, which equals a positive.

3. Expand 3z(z + a – 3)

Step 1 (write the base equation): 3z(z + a – 3)

Step 2 (this is a multiplication issue, so multiply everything in the brackets by ‘3z’): 3z2 + 3za – 9z

4. Expand then simplify the following equation: y(y - 3x) + x(3y - 2x)

Step 1 (writ e the base equation): y(y - 3x) + x(3y - 2x)

Step 2 (work from left to right, expanding the brackets): y2 – 3xy + 3xy – 2x2

Step 3 (cancel the like terms. Here -3xy and +3xy are like terms, so they cancel out): y2 – 2x2

Multiplying NegativesBasic rules:

Positive x Positive = Positive Negative x Negative = Positive Negative x Positive = Negative Positive x Negative = Negative Negative x Negative x Negative = Negative (remember BOMDAS) – e.g. -3 x -4 x -5 = 12 x -5 =

-60

Explanation about two negatives becoming positive from math is fun:

Examples

1. What is -3 x -5?

Step 1 (writ e the base equation): -3 x -5

Step 2 (remembering two negatives make a positive, do the basic arithmetic): 15

2. What is -4 x 7?

Step 1 (write the base equation): -4 x 7

Step 2 (remembering a positive and negative make a negative, do the basic arithmetic): -28

3. What is -7 2 ?

Step 1 (write the base equation): -72

Example: You owe 3 people $5 each. So you are "Negative 15" (3 × -$5 = -$15).

They then say "we like you so much we forgive the debt" ... you have just had 3 subtractions of -5, so it is like you have added $15 (-3 × -$5 = +$15).

You now have no debt: -$15 + $15 = $0

Step 2 (-72 is the same as writing -7 x -7, and since two negatives make a positive, we get): 49

4. What is -2 x 5 x -3?

Step 1 (write the base equation): -2 x 5 x -3

Step 2 (remembering BOMDAS, we work from left to right, so we do -2 x 5 first): -2 x 5 = -10.

Step 3 (the equation now becomes -10 x -3): 30

Commutative, Associative and Distributive LawsCommutative Laws let you swap the order of numbers and still get the same answer. This works for addition and multiplication but NOT division or subtraction.

Examples:

a + b = b + a; 6 +3 = 3 + 6 a x b = b x a; 4 x 3 = 3 x 4

Associative Laws let you change the grouping of numbers to still get the same answer. This works for addition and multiplication but NOT division or subtraction

Examples:

(a + b) + c = a + (b + c); (6 + 3) + 4 = 6 + (3 + 4) (a x b) x c = a x (b x c); (2 x 4) x 3 = 2 x (4 x 3)

Distributive Laws allow you to “distribute” the term outside the parentheses to the terms inside. Note this works for multiplication but NOT division

Examples:

a (b + c) = ab + ac; 4 (3 + 6) = 12 + 24

Examples

Cross Multiplication= ad = bc

Examples

1. ( x8 )=( 2x )Find x

Step 1 (write the base equation): ( x8 )=( 2x )

a =

c

b d

Step 2 (cross multiply): x2 = 8 x 2 = 16

Step 3 (solve. Remember that if we are wanting to find ‘x’ and we have ‘x2’ we need to isolate it. We do this by removing the exponent. We know that x2 = 16, so we do the exact opposite of what the exponent is doing by finding the square root of 16): sqrt(16) = 4 or -4.

2.

Step 1 (write the base equation): ( 5a )=( 1012 )Step 2 (cross multiply): 5 x 12 = 10a

Step 3 (isolate ‘a’ by removing the ‘10’): 60/10 = 6. Therefore, a = 6.

3.

Step 1 (write the base equation):

Step 2 (cross multiply): (x – 3)(x + 3) = 8 x 2 = 16

Step 3 (as we have two x’s and two 3’s, we need to find their difference. This is called the difference of two squares [subtract one from the other]): x2 – 32 = x2 – 9 = 16

Step 4 (we need to isolate ‘x’ so we first get rid of the 9 by adding it to 16): x2 = 25

Step 5 (remove the exponent): sqrt(25) = 5

Fractions in AlgebraNote that this method relies on following a couple of formulas listed below. The more common method would be to find the lowest common denominator.

Adding fractions

This method relies on finding the lowest common denominator. However, this formula isn’t very

common, but still works: ab+ cd=ad+bc

bd

For example: x+43

+ x−34

=(x+4 ) (4 )+(3 ) ( x−3 )

(3 ) (4 )=4 x+16+3 x−9

12=7 x+7

12

Subtracting fractions

This works the same way as adding, except we had a “-“ where the “+” is. ab− cd=ad−bc

bd

Multiplying Fractions

All we have to do is multiply the tops and bottoms together: abxcd= acbd

Dividing Fractions

When dividing fractions, “flip” the fraction you want to divide by.

ab÷cd=abxdc=adbc

Examples

1.

Step 1 (write the base equation): 2 yy−3

xy6

Step 2 (remember the formula for this kind of issue is : abxcd= acbd

so multiply the tops together and

bottoms together): 2 y x y6 ( y−3 ) =

2 y2

6 ( y−3 )= y2

3 ( y−3 )

2.

Step 1 (write the base equation): 2x−18

−3 x−212

Step 2 (apply the formula ab− cd=ad−bc

bd). 12 (2 x−1 )−8 (3 x−2 )

(8 ) (12 )=24 x−12−24 x−16

96

Step 3 (remove the like terms, which are 24x and -12 and -16): −12−−16

96= 496

= 124

3.

Step 1 (write the base equation): 8 x

3 ( y−1 )÷2 ( y−1 )6 x

Step 2 (apply the formula ab÷cd=abxdc=adbc

): 8 x x 6 x

3 ( y−1 ) x 2 ( y−1 ) = 48 x2

6 ( y+1 ) ( y−1 )

Step 3 (divide 48 by the 6 to simplify the denominator): 8 x2

( y+1 ) ( y−1 )

Step 4 (remembering the rule of ‘the difference of two squares, simplify the denominator further):

8 x2

y2−1

4.

Step 1 (write the base equation):

The more common method – Finding the lowest common denominator

Step 2 (apply the formula

ab÷cd=abxdc=adbc ):

What you take away from this one is that you don’t expand ‘x(y – 3)’ into ‘xy – 3x’. You should try and get rid of that x by dividing it into the numerator.

5.

Step 1 (write the base equation):

Step 2 (apply the relevant formula):

What you take away from this is not to expand ‘x(3 – x)’ into 3x – x2

6.

Step 1 (write the base equation):

Step 2 (start to apply the formula ab− cd=ad−bc

bd ):

Note here how the ‘2’ on the second denominator is applied to the term on the first denominator; that is, it becomes ‘2(y +2)’.

Step 3 (start to expand this part “ ”. It’s easier than you might think, apply each of the terms in the first/left brackets individually to the terms in the second/right brackets):

Note how applying the ‘y’ in the right brackets became 5y and also –y2 and so forth

Step 4 (start to simplify and remove terms):

And that is our answer

7.

Converting Repeating Decimals Into Fractions0.7… = 0.77777777777. The ellipsis means infinitely repeating. To turn this into a fraction, we need to introduce a variable like so: x = 0.7… = 0.777777. What would 10x be? It’d just be 10 x 0.7… = 7.777777… This is the trick, we know that x is 0.77777. The way we can get x is to subtract x from 10x because we will be left with a whole number. So:

10x = 7.7… AND we established x = 0.7… SO if we do 10x – x we have 9x (because we are subtracting one lot of x from 10 lots of x) and 7 (because 7.7... – 0.7… = 7. Putting this in a neater fashion:

1. 10x – x

2. 9x = 7. Solve for x.

3. x = 7/9

It is a similar method if the repeating decimals are of different numbers, for example 0.781… To solve this issue, you need to times by a factor of 10 so that only the repeating section of the decimal

will be to the right of the decimal point. For example :

10x = 34.1717171717… HOWEVER we cannot simply subtract x from 10x here because we would not be left with a whole number (note the.4 in x). SO we need to find 1000x because simply doing 100x would not give us a perfect repition of the decimals, it would be 341.71717

1000x = 3417.1717171717…

Subtract 10x from 1000x: 3417.171717 – 34.171717 = 3383. AND 1000x – 10x = 990x.

So our answer is 3383990

=3 413990

Examples

1. 1.2… Convert into a fraction

Step 1 (Set the problem equal to x): x = 1.2…

Step 2 (Multiply by 10): 10x = 12.2…

Step 3 (Subtract x from 10x): On the left hand side we get “10x – x = 9x” and on the right hand side we get “12.2… - 1.2… = 11”. So we have 9x = 11

Step 4 (isolate the x by dividing the right hand side by the left hand side): x = 11/9

Absolute Value EquationsAn absolute value equation would look something like this y = |2x – 1|. It’s in between the two vertical lines which essentially mean “convert to positive” OR “how far away from 0 is the equation” (this would give you the same answer either way, it’s just two different ways to think about it).

For example, |-1| = 1. -1’s distance from 0 is 1. OR you could just think “all I have to do is convert -1 into its positive number, which is 1.

Example: |x – 5| = 10. This is saying that the distance between x and 5 is 10. With a little bit of thought, we can see that this can lead us to two different answers. Either x – 5 = 10 OR x – 5 = -10. This is because we are finding the absolute value of the terms. We are saying that the absolute value of the equation (x – 5) is equal to 10. There is a possibility, however, that if we weren’t finding the absolute value, it would be equal to -10. So that’s why we have two different answer. So to solve, we would work through both possibilities:

1st possibility: x – 5 = 10

Add 5 to both sides: x = 15

-1 0 1

2nd possibility: x – 5 = -10

Add 5 to both sides: x = -5

So there’s two possible answers to this equation. Notice that both of these numbers are exactly 10 away from 5.

|4x – 1| = 19. 4x – 1 could be equal to 19 OR -19.

1st possibility: 4x – 1 = 19

Add 1 to both sides: 4x = 20

Divide both sides by 4: x = 5

2nd possibility: 4x – 1 = -19

Add 1 to both sides: 4x = -18

Divide both sides by 4 = -9/2

You can also graph these. Say you had y = |x + 3|. There’s two scenarios possible here.

1st scenario

The thing inside the absolute value is positive. Meaning y is a positive number, or you could just write it as x + 3 > 0 (if y = x + 3 and x + 3 is greater than 0, it follows that y must also be greater than 0). x + 3 is greater than 0 where x is greater than -3 (because if you added 3 onto -2 or any number greater, you’d get a positive). So where x > -3, the graph will look like y = x + 3.

2nd scenario

X + 3 < 0. In this equation, y = - (x + 3). So x has to be less than -3. So you’d get y = -x – 3 (which means Y would have to be a negative number).

The graph would look like this:

So the first scenario is the yellow, and the 2nd scenario is the purple slope. For the second slope, we can see that the y intercept is -3 and negative x means it slopes downwards, the x intercept would

be where y = 0. Y is equal to 0 where x = -3. For the first slope, the y intercept would be at 3, and the x intercept would be at -3 (because at -3, y would = 0)

Examples

1. Solve x: |2x – 5| = 11

Step 1 (find the two possibilities): the two possibilities are -11 and 11

Step 2 (work through each possibility. First possibility is -11): 2x – 5 = -11

Step 3 (add 5 to both sides): 2x = -6

Step 4 (divide both sides by 2): x = -3

Step 5 (work through the second possibility): 2x – 5 = 11

Step 6 (add 5 to both sides): 2x = 16

Step 7 (divide both sides by 2): x = 8

So the two answers are -3 and 8

2. 5|x + 3| - 3 = 7

Step 1 (only part of the equation is in absolute terms, so you want to isolate that part first): 5|x + 3| - 3 = 7

Step 2 (add 3 to both sides): 5|x + 3| = 10

Step 3 (divide both sides by 5): |x + 3| = 2 which means |x + 3| = 2 or – 2

Step 4 (work through each possibility by subtracting 3 from each side for each possibility): x = -1 OR -5 note that if you wanted to put these values back into the original equation, you would treat the absolute value sign as brackets and work through as you would with BOMDAS.

3. Solve |3x – 9| = 0 and graph the solution on a number line

Note here that there can be only one answer, which is 0. Add 9 to both sides of the equation 3x = 9. Divide by 3 you get x = 3. Now draw the solution on a number line. The solution is x = 3, so just draw a number line and draw a dot at 3.

4. Solve for x : 5|x – 8| - 2 = -6|x – 8| + 3

Step 1 (add 6|x-8| to both sides): 11|x – 8| - 2 = 3

Step 2 (add 2 to both sides): 11|x – 8| = 5

Step 3 (divide both sides by 11): |x – 8| = 5/11

Step 4 (write the two possibilities): |x – 8| = 5/11 OR -5/11

Step 5 (work through the first possibility): x – 8 = 5/11

Add 8 to both sides: x = 511

+8 = 511

+ 8811

=9311

Step 6 (work through the second possibility): x – 8 = -5/11

Add 8 to both sides: −511

+8 =- 511

+ 8811

=8311

5. Solve for x : 5|x + 7| - 5 = -5|x + 7| + 7

Step 1 (add 5|x + 7| to both sides): 10|x + 7| - 5 = 7

Step 2 (add 5 to both sides): 10|x + 7| = 12

Step 3 (divide both sides by 10): |x + 7| = 1210

=65

Step 4 (we now have the two possibilities, 6/5 OR -6/5, work through both): |x + 7| = 65

Subtract 7 from both sides: x = 65−355

=−295

Step 5 (work through the second possibility): |x + 7| = - 65

Subtract 7 from both sides:- 65−355

=−415

Absolute Values in InequalitiesStart with the example of |x| < 12. Remember an absolute value means how far away you are from 0. So, this issue is asking us, what are all the x’s that are less than 12 away from 0. This is easy – the numbers would simply be -11 to 11. Each number is less than 12 away from 0. So it could be all of the numbers where x > -12 and x < 12. If it meets both these constraints, it will have to be less than 12 away from 0.

|7 x|≥21 This means that whatever is in our absolute value sign must be 21 or more away from 0. So it would be either 7 x≤−21∨7 x ≥21 If it was equal to or less than 21 and you took it’s absolute value, then it’s going to be greater than or equal to 21. Divide each side by 7 you’d get. x≤−3∨x ≥3

Another example: |5x + 3| < 7. This is telling us that what ever is inside of the absolute value sign has to be less than 7 away from 0. You’d have to be somewhere in the range of -6 to 6. It would have to be expressed as such: 5 x+3>−7∧5x+3<7 in order for its absolute value to be less than 7. So for the first expression you’d subtract 3 and divide by 5 giving you x > -2 and for the second

expression you’d subtract 3 and divide by 5 giving you 45

. You’d express this as (-2, 45

¿ and this

means than x could be anything from -2 to 45

|2x7 +9|> 57This could mean that

2x7

+9> 57∨2x7

+9<−57

In the second possibility, if it’s less than -5/7 then

when you take the absolute value, it will have to be greater than 5/7.

First possibility: 2x7

+9> 57

Multiply both sides by 7: 2 x+63>5

Subtract 63 from both sides: 2x > -58

Divide by 2: x > -29

Second possibility: : 2x7

+9<−57

Multiply both sides by 7: 2 x+63←5

Subtract 63 from both sides: 2x < -68

Divide by 2: x < -34

Note this above example is essentially saying that the absolute value of y has to be negative. Therefore, there is no solution.

Factoring in Algebra + Special Binomial ProductsThese two sections have been combined because it is very difficult to factor without first knowing special binomial products.

Special Binomial Products

A binomial is a polynomial (haven’t covered this yet) with two terms:

A product means the result you get after multiplying. E.g. (a + b)(a – b) means (a + b) multiplied by (a – b).

Multiplying a binomial by itself

This refers to the situation where you are squaring a binomial. This is the formula:

(a + b)(a + b) = a2 + ab + ab +b2 = a2 + 2ab + b2 – this may seem complicated, but just think of it as apply all of the terms in the left brackets to those in the right brackets.

Subtract times subtract

This refers to the situation where you square a binomial with a minus symbol inside the brackets:

(a – b)(a – b) = a2 – ab – ab + b2 = a2 – 2ab + b2. Again, we are applying the terms in the left brackets to those in the right brackets. Notice we get ‘-ab’ twice as we would be applying a positive to a negative, which = a negative number.

Add times subtract

This refers to the situation where you multiply positive terms in one set of brackets by terms separated by a minus sign in another set of brackets i.e. (a + b)(a – b).

(a + b)(a – b) = a2 – ab + ab – b2 = a2 + 0 – b2 = a2 – b2. We get the ‘0’ because we have a negative ab plus a positive ab, which has to equal 0.

Summary

(a+b)2 = a2 + 2ab + b2 (a-b)2 = a2 - 2ab + b2

(a+b)(a-b) = a2 - b2 – we call this one the difference of two squares.

Examples

1. (y + 1) 2

Step 1 (write the base equation): (y + 1)2

Step 2 (apply the formula ‘(a + b)2 = a2 + 2ab + b2 ‘:(y+1)2 = (y)2 + 2(y)(1) + (1)2 = y2 + 2y + 1

2. (3x – 4) 2

Step 1 (write the base equation): (3x – 4)2

Step 2 (apply the formula (a-b)2 = a2 - 2ab + b2): (3x-4)2 = (3x)2 - 2(3x)(4) + (4)2 = 9x2 - 24x + 16

3. ( 4y+2)(4y-2 )

Step 1 (write the base equation): (4y+2)(4y-2)

Step 2 (apply the formula (a+b)(a-b) = a2 - b2): (4y+2)(4y-2) = (4y)2 - (2)2 = 16y2 - 4

4. 4x 2 – 9

Step 1 (write the base equation): 4x2 – 9

Step 2 (at the moment, this equation doesn’t seem to fit any of our formulas. But, we can apply one of these formulas if we find the right pattern. Here we have an exponent of 2, and both ‘4’ and ‘9’ have perfect square roots, so we can use this to apply the difference of two squares formula): 4x2 - 9 = (2x)2 - (3)2

Note we can achieve the same result by going through the difference of squares formula: (2x+3)(2x-3) = (2x)2 - (3)2 = 4x2 - 9

5. Factor 49x 2 - 42x + 9

Step 1 (write the base equation): 49x2 - 42x + 9

Step 2 (identify the relevant formula. Here, it looks like it would be(a-b)2 = a2 - 2ab + b2. This requires us to figure out the values of the numbers to substitute in place of the variables): a2 = 7x and b2 = 3. We can see if this checks out (7x)2 – 2(7x)(3) + 32 = 49x2 - 42x + 9.

Step 3 (put this into the form of (a – b)2 = (7x – 3)2

6. Factor 0.36x 2 - 6.25y 2

Step 1 (write the base equation): 0.36x2 - 6.25y2

Step 2 (apply the relevant formula here; (a+b)(a-b) = a2 - b2. You have to figure out the square root of 0.36x2 and 6.25y2): (0.6x + 2.5y)(0.6x - 2.5y)

7. Simplify (3x + 2y) 2 - (3x - 2y) 2

Step 1 (write the base equation): (3x + 2y)2 - (3x - 2y)2

Step 2 (work through the first lot of brackets, which requires the formula (a+b)2 = a2 + 2ab + b2):

(3x)2 + 2(3x)(2y) + (2y)2 = 9x2 + 12xy + 4y2

Step 3 (work through the second lot of brackets, which requires the formula (a-b)2 = a2 - 2ab + b2):

(3x)2 – 2(3x)(2y) + (2y)2 = 9x2 - 12xy + 4y2

Step 4 (subtract now): 9x2 + 12xy + 4y2 - (9x2 - 12xy + 4y2)

Step 5 (before we actually subtract, we should remove the brackets. This changes the values of the numbers in the brackets. For example the 9x2 becomes -9x2 and the -12xy becomes a positive 12xy): 9x2 + 12xy + 4y2 - 9x2 + 12xy - 4y2 = 24xy

FactoringNumbers have factors: e.g.

Expressions like x2 + 4x + 3 have factors:

So think of a ‘factor’ as a ‘factor’ (in the regular sense of the word) of the final equation.

Factoring is the process of finding the factors. It is like “splitting” an expression into a multiplication of simpler expressions. E.g. 2y + 6 can be factored into 2(y + 3). This works because ‘2y’ and ‘6’ have a common factor of 2.

Common factor

You need to make sure you have the highest common factor, including variables. For example, 3y2 + 12y = 3y(y + 4). When we expand ‘3y’ into the terms in the brackets, we get 3y2 + 12y.

More complicated factoring

For some of these, you will need to remember your special binomial products. For example if you needed to factor 4x2 – 9. Well, this could be broken down into (2x)2 – 32. But this looks a lot like our difference of two squares doesn’t it? Remember that (a+b)(a-b) = a2 - b2. So it can be broken down even further into (2x+3)(2x-3).

Factoring Identities (formulas)Expanded Form Factored Form

a2 - b2 (a+b)(a-b)a2 + 2ab + b2 (a+b)(a+b)a2 - 2ab + b2 (a-b)(a-b)

a3 + b3 (a+b)(a2-ab+b2)a3 - b3 (a-b)(a2+ab+b2)

a3+3a2b+3ab2+b3 (a+b)3

a3-3a2b+3ab2-b3 (a-b)3

Examples

1. Factor w 4 – 16

Step 1 (write the base equation): w4 – 16

Step 2 (look for common elements, here we could use an exponent of ‘2’): (w2)2 - 42

Step 3 (this is a familiar identity, that is, it is the difference of two squares. Apply this factoring identity): (w2 + 4)(w2 – 4).

Step 4 (w2 – 4 is another difference of two squares, so we need to factor that too. Note that w2 + 4 is ok to be left alone, as it is not any factoring identity we know): (w2 + 4)(w + 2)(w - 2)

2. Factor 3u 4   - 24uv 3

Step 1 (write the base equation): 3u4 - 24uv3

Step 2 (look for common elements. Here ‘3u’ could go into both 3u4 and ‘24u’. Remove this common element): 3u(u3 – 8v3)

Step 3 (the terms inside the brackets now correspond to a difference of cubes. Apply this formula to the terms inside the brackets ‘a3 - b3’): 3u(u3 - (2v)3) notice how ‘u3’ stays the same but we make ‘8v3’ into ‘(2v)3’

Step 4 (we’ve applied the expanded form of the difference of cubes, now we need to apply the factored form. If confused, see the table above, but we are essentially applying the following formula (a-b)(a2+ab+b2)): 3u(u-2v)(u2+2uv+4v2)

3. Factor z 3   - z 2   - 9z + 9

Step 1 (write the base equation): z3 - z2 - 9z + 9

Step 2 (factor the first two terms and second two terms individually): z2(z-1) - 9(z-1) note that

-9(z-1) is the same as the original -9z + 9 because -9 x -1 = +9

Step 3 (apply z2 to 9): (z2-9)(z-1)

Step 4 (z2-9 is the difference of squares, so apply that formula): (z-3)(z+3)(z-1)

4. Factor 2w 4 - 162

Step 1 (write the base equation): 2w4 – 162

Step 2 (find a common element. 2 is a common element): 2(w4 – 81)

Step 3 (find a common element to the terms in the brackets. We can use exponents of 2 on each): 2 (w2)2 – 92)

Step 4 (this is a difference of two squares, so we get): 2(w2 + 9)(w2 - 9)

Step 5 (w2 – 9 is also a difference of two squares): 2(w2 + 9)(w + 3)(w - 3)

5. Factor x 3 + 125

Step 1 (write the base equation): x3 + 125

Step 2 (find common elements. X has a cube exponent. The cube root of 125 = 5): x3 + 53

Step 3 (this is a sum of cubes issue, so apply a3 + b3 = (a + b)(a2 - ab + b2)): x3 + 53 = (x + 5)(x2 - 5x + 52) = (x + 5)(x2 - 5x + 25)

6. Factor 192z - 3z 4

Step 1 (write the base equation): 192z - 3z4

Step 2 (Find common elements. 3z is common to both, so remove the 3z): 3z(64 - z3)

Step 3 (64 has a perfect cube root, so find the cube root. Note that we remove the brackets): z3 = 43 - z3

Step 4 (this is a difference of cubes, so apply the formula a3 - b3 = (a - b)(a2 + ab + b2)): 43 - z3 = (4 - z)(42 + 4z + z2) = (4 - z)(16 + 4z + z2

7. 16y 2 -25x 2 +10x-1 – a trial and error issue

Step 1 (write the base equation): 16y2-25x2+10x-1

Step 2 (This issue is tricky, it does require trial and error. At this point, you could make two brackets based on the squares of 16 and 25): 4y(4y+5x) -5x(4y+5x) = 16y2 + 20xy -20xy -25x2 = 16y2 -25x2

Step 3 (we still need +10x-1, which we could get from adding +1 and -1 into the brackets): (4y-5x+1)(4y+5x-1) = 4y(4y+5x-1) -5x(4y+5x-1) +1(4y+5x-1)

= 16y2 + 20xy - 4y - 20xy -25x2 +5x + 4y +5x -1= 16y2 -25x2 +5x +5x -1 = 16y2 -25x2 +10x -1

8. Factor a 5 + a 2 b 3 - a 3 b 2 - b 5

Step 1 (write the base equation): a5 + a2b3 - a3b2 - b5

Step 2 (factor the first two terms. If we use ‘a2’ we can leave both terms inside the brackets as cubes): a5 + a2b3 = a2(a3 + b3)

Step 3 (factor the last two terms. If we use b2 we can leave both terms inside the brackets as cubes): - a3b2 - b5 = - b2(a3 + b3)

Step 4 (so at the moment we have a2(a3 + b3) - b2(a3 + b3). As it happens, you can simplify this further into): (a2 - b2)(a3 + b3) I don’t know why, but the math works out

Step 5 (note that the first bracket is a difference of two squares): (a + b)(a - b) (a3 + b3)

Step 6 (notice that the last bracket is a sum of two cubes): (a + b)2(a - b)(a2 - ab + b2)

9. Factor a 6 - b 6

Step 1 (write the base equation): a6 - b6

Step 2 (reduce the exponents into something more workable, like squares): (a3)2 - (b3)2

Step 3 (this gives us a difference of two squares problem): (a3 + b3)(a3 - b3)

Step 4 (the first bracket is a sum of two cubes issue): a3 + b3 = (a + b)(a2 - ab + b2)

Step 5 (the second lot of brackets from step 3 is a difference of two cubes issue): (a - b)(a2 + ab + b2)

Step 6 (put this together): a6 - b6 = (a + b)(a2 - ab + b2)(a - b)(a2 + ab + b2)

Introduction to LogarithmsLogarithms ask and answer the question “how many of one number do we multiply to get another number”. As such, it is closely related to exponents.

Example: How many 3s do we multiply to get 27? 3 x 3 x 3 = 27, so we multiply 3 3’s to get 27. The logarithm is 3, and we write it as. Log3(27) = 3

The base is the number we are multiplying. For example, we could say:

"the logarithm of 8 with base 2 is 3" or "log base 2 of 8 is 3" or "the base-2 log of 8 is 3"

As already mentioned, logarithms share a relationship with exponents. We can see why below:

23 = 8 is the same as log2(8) = 3

Common Logarithms: Base 10

note this button is the logarithm button on windows OS calculator default program. Note, however, that it has a base of 10. So every logarithm entered is essentially log10(x)

Natural Logarithms: Base “e”

We already mentioned that a base of 10 is natural, but mathematicians often use ‘Euler’s Number’ (‘e’) which is approx. 2.71828 as the base log. On a calculator it is this button:

Example: ln(7.389) = loge(7.389) ≈ 2. Because 2.718282 ≈ 7.389

Note that these common or natural bases apply where there is no base specified i.e. log3 would not be log3x10

Negative Logarithms

Negative logarithms indicate how many times to divide by the number. Example: log8(0.125) = -1.

Examples

1. What is log 4(256)?

Step 1 (write the base equation): log4(256):

Step 2 (figure out the exponent to apply): 44 = 256 so log4(256) = 4.

2. What is log 5(0.0016)?

Step 1 (write the base equation): log5(0.0016)?

Step 2 (apply negative exponents): 5-4 = 0.0016. So log5(0.0016) = -4

3. Write log 3(2,187) = 7 in exponential form

37 = 2,187

4. What is the value of log e(5)

Press ‘5’ then ‘In’ on the windows calculator: 1.609

PolynomialsPolynomial means “many terms”. It can have:

Constants: like 3, -20, or ½ Variables: (like x and y) Exponents: like y2, y0 But note that negative exponents and fractional exponents are NOT

polynomial compatible. Example: 3xy-2 is NOT a polynomial

Note also polynomials can be combined using addition, subtraction, multiplication and division EXCEPT not division by a variable.

Examples of polynomials:

3x x - 2 -6y2 - (7/9)x 3xyz + 3xy2z - 0.1xz - 200y + 0.5 512v5+ 99w5

1 x/2 is allowed, because you can divide by a constant also 3x/8 for the same reason √2 is allowed, because it is a constant (= 1.4142...etc)

Examples of what are not polynomials:

2/(x+2) is not, because dividing by a variable is not allowed 1/x is not either 3xy-2 is not, because the exponent is "-2" (exponents can only be 0,1,2,...) √x is not, because the exponent is "½"

Degrees and Standard Form

The degree of a polynomial with only one variable is the largest exponent of that variable. Example:

4x3 – x + 3. The Degree is 3 (largest exponent of x)

The standard form for writing a polynomial is to put the terms with the highest degree first. Example, put the following into standard form:

3x2 - 7 + 4x3 + x6 = x6 + 4x3 + 3x2 – 7

Examples

1. What is the degree of the polynomial 5x 3 - 8x + 3x 5 + 4x 2 - 7x 4 + 1?

Put this into standard form for writing a polynomial: 5x3 - 8x + 3x5 + 4x2 - 7x4 + 1 = 3x5 - 7x4 + 5x3 + 4x2 - 8x + 1

2. What is the degree of the polynomial 3x 4 - 4x 3 y + 6x 2 y 2 - 8x 2 y 3 - 7y 4 + 2?

Calculate the total degrees for each term. You’ll find that 8x2y3 is 2 + 3 = 5

Adding and Subtracting PolynomialsTo add polynomials, add any like terms together.

Like terms are terms whose variables and exponents are the same. The coefficients like “5” in 5x can be different

Adding polynomials

There are two steps to follow here:

1. Place the like terms together2. Add the like terms

For example, add 2x2 + 6x + 5 and 3x2 - 2x – 1

Step 1 (write the base equation):

Step 2 (identify the like terms and place them together): 2x2 + 3x2 + 6x – 2x + 5 – 1

Step 3 (add the like terms): (2 + 3)x2 + (6 – 2)x + (5 – 1) = 5x2 + 4x + 4

A helpful way to add several polynomials is to use the column method. For example:

Add (2x2 + 6y + 3xy), (3x2 - 5xy - x) and (6xy + 5)

Subtracting Polynomials

To subtract polynomials, reverse the sign of each term you are subtracting (e.g. “+” turns into “-“ and “-“ into “+”), then add as usual. For example:

5y2 + 2xy – 9 – (2y2 + 2xy – 3)

Step 1 (write the base equation): 5y2 + 2xy – 9 – (2y2 + 2xy – 3)

Step 2 (reverse the signs of the terms in the brackets): 5y2 + 2xy – 9 – 2y2 - 2xy + 3

Step 3 (do the arithmetic): 3y2 – 6

Examples

1. Add the polynomials (3x 2 - 6x + xy), (2x 3 - 5x 2 -3y) and (7x + 8y)

Step 1(Write the base equation):

That is our answer, but note the order of those terms in the answer doesn’t matter, so long as we have the right symbol before them i.e. + and –

2. Subtract (-3x2 + 5y - 4xy + y²) from (2x2 - 4y + 7xy - 6y²)

Step 1 (write the base equation): Subtract (-3x2 + 5y - 4xy + y²) from (2x2 - 4y + 7xy - 6y²)

Step 2 (write the equation the right way around. Notice that it says “from” in the problem): (2x2 - 4y + 7xy - 6y2) - (-3x² + 5y - 4xy + y2)

Step 3 (change the signs): 2x2 - 4y + 7xy - 6y2 + 3x2 - 5y + 4xy - y2

Step 4 (add the like terms): 2x2 + 3x2 - 4y - 5y + 7xy + 4xy - 6y2 - y2

Step 5 (do the arithmetic): 5x2 - 9y + 11xy - 7y2

3. Subtract (4x2 + 2x - 3y + 7xy - 3y2) from (3x2 - 5x + 8y + 7xy + 2y2)

Step 1 (write the base equation the right way around): (3x2 - 5x + 8y + 7xy + 2y2) - (4x2 + 2x - 3y + 7xy - 3y2)

Step 2 (change the signs around): 3x2 - 5x + 8y + 7xy + 2y2 - 4x2 - 2x + 3y - 7xy + 3y2

Step 3 (bring the like terms together): 3x2 - 4x2 - 5x - 2x + 8y + 3y + 7xy - 7xy + 2y2 + 3y2

Step 4 (combine like terms): -x2 - 7x + 11y + 5y2

4. If P = 4x 4 - 3x 3 + x 2 - 5x + 11 and Q = -3x 4 + 6x 3 - 8x 2 + 4x – 3 what is 2P + Q?

Step 1 (write the base equation): 2(4x4 - 3x3 + x2 - 5x + 11) + -3x4 + 6x3 - 8x2 + 4x – 3

Step 2 (figure out ‘P’): 8x4 - 6x3 + 2x2 - 10x + 22

Step 3 (add to ‘Q’): 8x4 - 6x3 + 2x2 - 10x + 22 + (-3x4 + 6x3 - 8x2 + 4x - 3)

Step 4 (combine the like terms): 8x4 - 3x4 - 6x3 + 6x3 + 2x2 - 8x2 - 10x + 4x + 22 - 3

Step 5 (do the arithmetic): 5x4 - 6x2 - 6x + 19

5. Add the polynomials (-4x 3 + x 2 y - xy 2 ), (3x 3 - xy 2 + 5x 2 y) and (7xy 2 + 3y 3 )

Step 1 (write the base equation): (-4x3 + x2y - xy2) + (3x3 - xy2 + 5x2y) + (7xy2 + 3y3)

Step 2 (do the column method):

Note also that the order of the terms doesn’t matter so long as the correct symbols are used, so these answers are also correct: -x3 + 5xy2 + 6x2y + 3y3 AND -x3 + 3y3 + 6x2y + 5xy2

6. Add the polynomials (5x2y - xy2+ 5x - 3), (4xy2 - 3y + 7xy - 5) and (2x2y - 3xy).

7. If P = 3x2 + xy - 5y2, Q = 2x2 - xy + 3y2 and R = -6x2 + 4xy - 7y2, what is P + Q - R?

Step 1 (write the base equation): (3x2 + xy - 5y2) + (2x2 - xy + 3y2) - (-6x2 + 4xy - 7y2)

Step 2 (change the direction of the symbols in the last bracket): 3x2 + xy - 5y2 + 2x2 - xy + 3y2 + 6x2 - 4xy + 7y2

Step 3 (combine the like terms): 3x2 + 2x2 + 6x2 + xy - xy - 4xy - 5y2 + 3y2 + 7y2

Step 4 (do the arithmetic): 11x2 - 4xy + 5y2

Multiplying PolynomialsMultiply each term in one polynomial by each term in the other polynomial and add those answers together, and simplify if needed.

1 term x 1 term (mononomial times mononomial)

Multiply the constants, then multiply each variable together and combine the result. Example:

(2xy)(4y) = 8xy2

1 term x 2 terms (monomial times binomial)

Multiply the single term by each of the two terms. Example:

2x (x + 3xy) = 2x2 + 6x2y

2 terms x 1 term (binomial times mononomial)

Multiply each of the two terms by the single term. Example:

(x2 – x)3y = 3x2y – 3xy

2 terms x 2 terms (binomials times binomial)

Each of the two terms in the first binomial is multiplied by each of the two ters in the second binomial, giving us 4 different multiplications. Example:

(a + b)(c + d) = ac + ad + bc + bd

The mathisfun website has some handy ways of remembering it here: http://www.mathsisfun.com/algebra/polynomials-multiplying.html

Firsts: ac

Outers: ad

Inners: bc

Lasta: bd

Another example:

(2x + 3)(xz – a) = 2x2z – 2xa + 3xz – 3a

2 terms x 3 terms (binomial times trinomial)

The FOIL method doesn’t work here. We have to multiply each term in the first polynomial by each term in the second polynomial. Example:

(x + a)(2x + 3y – 5) = 2x2 + 3xy – 5x + 2ax + 3ay – 5a

Like Terms

If we have like terms, remember to add them. Example:

(x + 2y)(3x – 4y + 5)

= 3x2 – 4xy + 5x + 6xy – 8y2 + 10y

= 3x2 + 2xy + 5x – 8y2 + 10y

Examples

1. Multiply out (3x + 2)(4x – 5)

Step 1 (write the base equation): (3x + 2)(4x – 5)

Step 2 (multiply the terms in the first lot of brackets by those in the second lot of brackets): 12x2 – 15x + 8x – 10

Step 3 (add the like terms, here it is ‘-15x’ and ‘8x’): 12x2 – 7x – 10

2. Multiply out (x – 4)(3x – y + 3)

Step 1 (write the base equation): (x – 4)(3x – y + 3)

Step 2 (multiply the terms in the first lot of brackets by those in the second lot of brackets): 3x2 – xy + 3x – 12x + 4y – 12

Step 3 (add the like terms. Here they are ‘3x’ and ‘-12x’): 3x² - xy - 9x + 4y - 12

3. What is the product of (3x – 4y) and (-2x + 5y – 6)

Step 1 (write the base equation): (3x – 4y)(-2x + 5y – 6)Step 2 (multiply the terms in the first lot of brackets by those in the second lot of brackets): -6x2 + 15xy - 18x + 8xy - 20y2 + 24y

Step 3 (add the like terms, here they are ‘15xy’ and ‘8xy’): -6x2 + 23xy - 18x - 20y2 + 24y

4. What is the product of (2y - 1), (2y + 1) and (4y 2 + 1)?

Step 1 (write the base equation): (2y - 1)(2y + 1)(4y2 + 1)?

Step 2 (BOMDAS tells us to work through the first two brackets first, so multiply those two together): 4y2 + 2y - 2y – 1 = 4y2 – 1

Step 3 (multiply this answer by the third bracket): (4y2 - 1)(4y2 + 1) = 16y4 + 4y2 - 4y2 – 1

Step 4 (simplify by adding like terms together. Here they are ‘4y2’ and ‘-4y2’): 16y4 – 1

5. What is the product of (3x - 2), (2x + 3) and (2x - 3)?

Step 1 (write the base equation. Note that there are two ways of doing this. See the alternative below): We will write the equation as (3x - 2)[(2x + 3)(2x - 3)]

Step 2 (apply the difference of two squares formula to the terms in the square brackets): (3x - 2)(4x2 - 9)

Step 3 (multiply the brackets together): 12x3 - 27x - 8x2 + 18

Step 4 (rearrange in order of degree): 12x3 - 8x2 - 27x + 18

And the alternative way

(3x – 2)(2x – 3) = 6x2 + 9x – 4x -6 = 6x2 + 5x – 6

(6x2 + 5x – 6)(2x – 3) = 12x3 – 18x2 + 10x2 – 15x - 12x + 18 = 123 – 8x2 – 27x – 18

6. What is the product of (2y 2 - 9), (y + 3) and (y - 3)?

Step 1 (write the base equation): (2y2 - 9) [(y + 3)(y - 3)]?

Step 2: (2y2 - 9)(y2 - 9)

Step 3: 2y4 - 18y2 - 9y2 + 81

Step 4: 2y4 - 27y2 + 81

Long Polynomial MultiplicationWhen we are faced with polynomials involving 3 or more terms it can be easier to use a colum method. This works in a similar way to if we were just doing basic long multiplication without any variables. This is the method:

1. Multiply the polynomial (usually the longest one) by the first term of the other polynomial2. Repeat for the second term of the other polynomial3. Keep repeating this process4. Add up columns

REMEMBER to always make sure you line up the like terms in the same columns. See the below example, the like terms with ‘x’ are lined up, and the like terms with no variables are lined up too.

Example:

The yellow arrows signify the first step, i.e. multiply the longest polynomial (in this case, it is ‘(3x2 + 2x – 8)) by the first term of the other polynomial. The first term here is ‘5’, so we are multiplying everything in the first lot of brackets by 5. The purple arrows signify the second step, which is multiplying the longest polynomial by the second term in the other polynomial, so everything is being multiplied by ‘x’.

What if one polynomial has an ‘x2’ but the other doesn’t? easy, you just leave gaps. Example:

More than one variable

So far we’ve seen one variable in our issues (‘x’). What if there’s more than one? Simply write down the answers as they come, but always check to see if your answer could fit under a matching answer. Example:

Examples

1. Multiply out (4x² - 3x + 5)(x - 4) using polynomial long multiplication.

2. Multiply out (3x 3 + 4x - 5)(2x 3 - 8x 2 + 2) using polynomial long multiplication.

3. Multiply out (5x 3 + 2x - 7)(2x - 3) using polynomial long multiplication.

4. Multiply out (x 2 + 3xy - y)(3x + y) using polynomial long multiplication.

5. Multiply out (3y 2 + 5y - 7)(y + 4) using polynomial long multiplication.

6. Multiply out (-3x 4 + x 3 - 7x)(2x 2 - x + 4) using polynomial long multiplication.

7. Multiply out (x 3 - 3xz + z 2 )(x - z) using polynomial long multiplication.

8. Multiply out (2x 2 - 5x + 6)( 4x 2 + 3x - 8) using polynomial long multiplication.

9. Multiply out (3 - 5y + y 2 )(15 + 7y 2 - 2y 3 ) using polynomial long multiplication.

Dividing polynomialsThere are a number of different methods that we can use to divide a polynomial.

First method: splitting the “+” and “-“ signs

6 x−33

=6 x3

−33=2 x−1

Note that we keep the denominator for both parts of our fraction. We split the positive 6x and the -3 and divided each by the same denominator.

Another longer example is:

Note that the x2 just became x. That’s because it was being divided by an x. Further, we get 1/3x because we can’t simplify that any further.

Second method: rearranging top polynomial

2x2 + 2x is the same as 2x(x + 1), so that’s where that comes from.

Third method: long division

The denominator goes first, then follows the numerator. Note that they should always be placed in order of degree (that is, put the highest exponents first).

There are some steps to work through here. Take the following equation for example:

Step 1 (divide the x2 by the ‘x’, put the answer above the line. We will get just 1 x):

Step 2 (multiply the answer we just got i.e. x by the denominator, i.e. x + 2. This time, write it under

our equation. x(x + 2) = x2 + 2x):

Step 3 (we now do subtraction. Subtract our previous answer from the numerator. Note we subtract

downwards in the columns):

Step 4 (repeat this entire process. So divide the -5x by the ‘x’ in the denominator. This will just give

us 5, as we are essentially removing the x):

Step 5 (because we are repeating the process, we are back at the multiplication stage. So multiply the ‘-5’ by x + 2. Which will give us -5x and – 10):

Step 6 (subtract downwards again):

And, to check your answer, you can multiply it by the bottom polynomial. This should give you the

top polynomial:

Long division - Remainders

So far we’ve been looking at situations in long division in which numbers divide perfectly into one another. Sometimes they don’t. Let’s look at an example:

Step 1 (divide ‘2x2’ by ‘x’):

Step 2 (multiply 2x by ‘x – 3’):

Step 3 (subtract downwards):

Step 4 (divide ‘1x’ by ‘x’):

Step 5 (multiply ‘1’ by ‘x – 3):

Step 6 (subtract. This is where we get our remainder):

The remainder came from the 0 + 2 at the bottom, which is just ‘2’. Write that over the denominator.

Long Division with “Missing” Terms

Sometimes you’ll get an issue with “missing terms”. An example of this would be if you had an x4 but no x2. To overcome this, include the missing terms with a coefficient of 0 to give you enough space. That may not make sense right now, but let’s do an example. I’m not going to say what we’re doing at each step. We are doing the steps we did above, only now we’re leaving blanks. Each box is the next step in the process of solving this issue. It is for you to remember what we are doing to get there:

We can see that the reason we use the coefficient of 0 to make up missing terms is that it gives us the necessary space to complete the problem without interfering with our calculations.

Long Division with More than One Variable

The issues we’ve seen have only had ‘x’ as the variable. What if x and y were there? The method is still the same. Same as above, I’ll give the issue and the steps we take in figuring it out, but I won’t write why:

Examples

1. Divide 8x 2 + 12x - 3 by 4x

Step 1 (write the base equation):

Step 2 (split the positive and negative signs):

Step 3 (simplify):

2. Divide (5x 2 - 10x) by (x - 2), where x ≠ 2

Step 1 (write the base equation):

Step 2 (factor the numerator):

Note that x cannot = 2 otherwise we’d be dividing 0 by 0 which is undefined.

3. Divide (12x 3 + 9x 2 - 3) by 3x

Step 1 (write the base equation):

Step 2 (split the positives and negatives):

Step 3 (divide):

4. Divide (-12x 2 - 18x) by (2x + 3)

Step 1 (write the base equation):

Step 2 (factor the numerator): = -6.

Note that x cannot = -3/2 otherwise we would be dividing 0 by 0 which is undefined.

5. Divide (6x 3 - 2x 2 - 15x + 5) by (3x - 1)

Step 1 (write the base equation):

Step 2 (look for common factors to factor the numerator): =

Note that x cannot = 1/3 otherwise we would be dividing 0 by 0, which is undefined.

6. Divide (a 3 + b 3 ) by (a + b)

Step 1 (write the base equation):

Step 2 (factor the numerator. a3 + b3 is a sum of cubes which = (a+b)(a2-ab+b2)):

7. Divide (x 2 + 2x -15) by (x + 5), where x ≠ -5

8. Divide (x 3 - y 3 ) by (x - y)

Step 1 (write the base equation):

Step 2 (factor the numerator. This is a difference of cubes which we know factors into (x - y)(x2 + xy +

y2)): = x2 + xy + y2

9. Divide (8p 3 + 27q 3 ) by (2p + 3q)

Step 1 (write the base equation):

Step 2 (write this as a sum of cubes issue = (a+b)(a2-ab+b2):

10. Divide (2x 5 - 5x 4 + 7x 3 +4x 2 - 10x + 11) by (x 3 + 2)

11. Divide the polynomial (3x 3 - 11x 2 y + 11xy 2 - 2y 3 ) by the binomial (x - 2y)

Note that we get -5x2y because we are subtracting a negative from a negative, so we don’t get -17

12. Divide (6x 2 - 17x + 12) by (3x - 4)

13. Divide (10x 5 + x 3 + 5x 2 - 2x - 2 ) by (5x 2 – 2)

Note that we get +5x3 because x - -4x = 5x

14. What is the remainder when (2y 3 - y 2 - 13y + 9) is divided by (y - 2)?

15. What is the remainder when (5x 3 + 3x 2 + 8x - 8) is divided by (5x - 2)?

16. What is the remainder when (6x 3 - 13x 2 - 4x + 35) is divided by (3x + 4)?

17. What is the remainder when (5x 4 - 13x 3 + 21x 2 + x + 10) is divided by (5x - 3)?

18. What is the remainder when (6x 4 + 11x 3 - 7x 2 - 15x - 50) is divided by (3x + 7)?

ConjugateThe conjugate is where the change middle sign of two terms, for example:

Expression Its ConjugateX2 – 3 X2 + 3a + b a – ba – b3 a + b3

Knowing conjugates can be helpful. For example, because of our knowledge of these conjugates, we know that when you multiply something by its conjugate you get squares like this:

Further, it can help us to remove a square root from the bottom or top of a fraction. Note that if we remove an irrational expression from the denominator, we are said to be “rationalizing the denominator”.

For example, the following has an irrational denominator:

We want to move the square root of 2 to the top. We do this by multiplying both top and bottom by the conjugate (this won’t change the value of the fraction):

Note: the denominator became a2 – b2

Examples

1. What is the result of multiplying (3 - 2x) by its conjugate?

Step 1 (write the base equation): (3 – 2x)(3 + 2x).

Step 2 (multiply by multiplying the terms in the left hand brackets to those in the right): 3(3 + 2x) - 2x(3 + 2x)

= 9 + 6x - 6x - 4x2

= 9 - 4x2

2. What is the result of adding (5y + 4) to its conjugate?

Step 1 (write the base equation): (5y + 4) + (5y – 4)

Step 2 (add together, adding like terms): 10y

3. What is the result of multiplying (5x 2 + 3x) by its conjugate?

Step 1 (write the base equation): (5x2 + 3x)(5x2 - 3x)

Step 2 (apply the terms in the left brackets to those in the right): 5x2(5x2 - 3x) + 3x(5x2 - 3x)

= 25x4 - 15x3 + 15x3 - 9x2

= 25x4 - 9x2

Note we could just use the difference of two squares formula: (5x2 + 3x)(5x2 - 3x) = (5x2)2 - (3x)2 = 25x4 - 9x2

4. What is the result of subtracting (3x 2 - 2x) from its conjugate?

Step 1 (write the base equation): (3x2 + 2x) - (3x2 - 2x)

Step 2 (remove the brackets. This will change the signs around): 3x2 + 2x - 3x2 + 2x = 4x

5.

Step 1 (rewrite the base equation, including the conjugate):

Step 2 (apply the difference of squares to the denominator):

Step 3 (work out the denominator in terms of whole numbers): =

Step 4 (simplify this by dividing both top and bottom by 2):

6.

Step 1 (write the base equation, including the conjugate):

Step 2 (apply the difference of squares to the denominator):

Step 3 (work out what the denominator is equal to):

Step 4 (simplify. We can divide numerator and denominator by 3):

7.

Note this one is a bit of a trick. We’re being asked to find the conjugate of the rational expression in the question. So you have to rationalise the denominator like always, but then flip the signs.

Step 1 (write the base equation, including its conjugate):

Step 2 (apply difference of squares):

Step 3 (work out the denominator): =

Step 4 (we can’t simplify into a whole number, but we can split into individual fractions):

Step 5 (this is where you flip the signs):

8.

Now we can find the conjugate by changing the sign in the middle:

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

Linear Equations

Equation of a straight lineThe equation of a straight line is written as

y=mx+b OR y=mx+c

Where ‘m’ is the variable representing the slope or ‘gradient’ (literally how steep the line is), ’x’ representing how far along the line is, and ‘b’ representing the Y Intercept (that is, where the line crosses the Y axis).

Finding ‘m’ and ‘b’

To find ‘b’, just look where the line crosses the Y axis.

To find ‘m’ (the slope), the formula is as follows: Change∈YChange∈X

sometimes also expressed as m= riserun

In this example, to calculate ‘m’, we get ‘m’ = 2/1 = 2. We can see that ‘b’ = 1 because that’s where it crosses the Y-axis. Therefore y = 2x + 1. Now, we can assign any value we like to ‘x’ and get the corresponding value for ‘y’. For example, if we assigned ‘x’ the value of 1, then we get y = 2 x 1 + 1 = 3. Therefore y = 3 when x = 1. Look up at the graph, we can see that when y = 3, x = 1.

We could choose another value for x, such as 7. y = 2×7 + 1 = 15. And so when x=7 you will have y=15

But how did you know to choose 1 and 2? They seem like pretty arbitrary numbers. You’re right, they are. You could have chosen to extend that red line out to 1.2, gone directly up to where it intersects the line and you would have ended up at 2.4, and 2.4/1.2 = 2.

If we have a purely vertical line, then the slope is undefined, and does not cross the Y-Axis. The formula changes, we use “x = …”. You will need to look to the graph to determine this: Example:

x = 1.5. Every point on the line has x coordinate 1.5

Examples

1. For the straight line y = -2x + 3 , what are: a) the slope b) the y-intercept?

Slope = -2. Y-Intercept = 3

2. What is the equation of the straight line shown in the diagram?

Step 1 (find the slope or gradient): Slope = change in y / change in x = 2/4 = ½. Therefore, m = ½

Step 2 (find the y-intercept by looking at the graph): Y-Intercept = -2. So b = -2.

Step 3 (apply what we know to the base formula y = mx + b): y = (1/2)x – 2

3. What is the equation of the following line?

Step 1 (calculate the slope): m = rise/run = ¾ = 0.75

Step 2 (find b): Y-Intercept = 2

Step 3 (substitute into y = mx + b): y = 0.75x + 2

4. What is the equation of the following graph?

Step 1 (calculate the slope): m = rise/run = -4/1 = -4

Step 2 (find the Y-Intercept): b = 2

Step 3 (substitute into y = mx + b): y = -4x + 2

5. For the straight line x = 2y - 3 , what are: a) the slope b) the y-intercept?

Step 1 (start to rearrange. Note that we currently have x = 2y – 3, so we need to isolate ‘y’): x + 3 = 2y; x(1/2) + 1½.

Step 2 (substitute into y = mx + b): we can now see that slope = ½ and y intercept = 1½

6. What is the equation of the straight line shown in the diagram?

Step 1 (find the slope): (change in y) / (change in x) = -4/3.

Step 2 (find the y-intercept): -4. So b = -4

Step 3 (substitute into the equation y = mx + b):

7. What is the equation of the straight line shown in the diagram?

Step 1 (find the slope): (change in y) / (change in x) = 6/2 = 3

Step 2 (find the Y-Intercept): -5

Step 3 (apply to the formula y = mx + b): y = 3x – 5

8. What is the equation of the straight line shown in the diagram?

Step 1 (find the slope): (change in y) / (change in x) = -3/5 = -0.6

Step 2 (find the Y-Intercept): 2.5

Step 3 (substitute into the equation y = mx + b): y = -0.6x + 2.5

Exploring the Line GraphHere, we can see the slope is negative (negative 1) and now it’s sloping to the left.

Here, I changed the slope to being positive (m = 2), and we can now see it is sloping right.

Now, I went back to the slope = -1, but changed the value of ‘b’ to -1.9. We can see here that it moved downwards to the left. If we made it positive, it would move upwards to the right.

Now, I changed the slop back to being positive, keeping a negative value for ‘b’. We can see that it also moved downwards, but downwards here means going down to the right.

Cartesian CoordinatesCartesian coordinates help pinpoint where certain things are on maps/graphs.

They are expressed like ‘(6, 5)’. This means 6 points across from 0 on the x-axis and 5 up on the y-axis.

X axis Horizontal

Y axis Vertical

Axis: The reference line from which distances are measured.. The plural of Axis is Axes, and is pronounced ax-eez

Example of plotting Cartesian coordinates (6, 4):

Examples using negatives:

And then there’s quadrants. There are 4 (hence ‘quad’):

3 dimensional coordinates

To see why we get these coordinates, see here:

Examples

1. What are the Cartesian coordinates of the point A?

Answer: (4, 2) – 4 across and 2 up

2. What are the Cartesian coordinates of the point B?

Answer: (-3, 5) = -3 back, 5 up

3. In which quadrant do you find the point (3, -2)?

Answer: fourth quadrant:

3. What are the three dimensional coordinates of the point P?

ExplanationAnswer: (5, 4, -2). Explanation: This is what I think is the rationale behind getting these coordinates. Mathisfun didn’t actually tell us why. I’d personally have picked (6, 5, -8), that just seems to me where those points are. But with a bit of thought, we can see that this is wrong. Keeping in mind that this is a representation of a 3 dimensional object, so there is the issue of depth to deal with. Let’s

start with how we got the ‘5’ on the x-axis. See how the vertices (and also edges) touch x coordinate 5 exactly. Well, think about that, point P is directly behind this point, only it is a bit higher, so it is still 5 across. So, we get to the Y-Coordinate ‘4’. Look at point ‘4 ‘ on the Y-Axis where the horizontal edge touches it. Just like before, picture this shape in your head; that point (4) is exactly as high as point P, they are on the same plane, the difference is that point P is further across (that’s where we get point 5 from) and a bit behind (this is the issue of depth). So we get to the -2. Look at the Z-Axis. We can see that the numbers ascend as they go down to the left. So we move backwards to where the edge/vertice of the shape lines up with one of the lines. We count back 2 and see one of the back edges touches it. This back edge is exactly as deep as point P, so it has the coordinate of -2. Putting this all together, we get (5, 4, -2). Easy.

4. A = (-3, 3), B = (5, 3), C = (2, -2) and D = (-6, -2) . A is joined to B, B to C, C to D and D to A with straight lines. What shape is ABCD?

Answer: a parallelogram

Rational ExpressionsAn expression is the ratio of two polynomials. Think of it as a fraction with polynomials:

Examples of what is and what is not a rational expression

What is:

1

1−x2The top polynomial is “1” which is fine to have

2 x2+3 This is fine because we can write this as 2x2+31

What is not:

2−√ x4−x

The top is not a polynomial (square roots are not allowed)

1−x

1+( 1x ) 1/x is not allowed in a polynomial.

Generally

A rational function is the ratio of two polynomials P(x) and Q(x) like this

note: Q(x) can’t be 0, and if it is, it is undefined

Finding Roots of Rational Expressions

A root is where the expression is equal to. To find the roots of a rational expression you only need to find the roots of the top polynomial, so long as the Rational Expression is in “Lowest Terms”.

Lowest terms

A fraction is in lowest terms with the top and bottom have no common factors. Example: 26=13

Likewise, for rational expressions

Solving PolynomialsTo solve a polynomial means to find the “roots”. A “root” is where the function is equal to 0. Solving depends on the degree.

The degree of a polynomial with one variable is the largest exponent of that variable.

How to solve?

Quadratic EquationsThe name Quadratic comes from “quad” meaning square, because the variable gets squared:

The standard form of a quadratic equation looks like this:

A, b, and c are known values; a can’t be 0

Examples of quadratics:

= In this one a=2, b=5 and c=3

a = 1, but we don’t usually write “1x2” b = -3

c = 0, so we don’t show that

= not a quadratic, as it is missing x2

Sometimes a quadratic is hidden, and you need to do a bit of work to an equation to find out that it is indeed a quadratic.

The Golden Ruleax2 + bx + c = 0

In Disguise In Standard Form a, b, and cx2 = 3x -1 Move all terms to

left hand sidex2 - 3x + 1 = 0 a=1, b=-3, c=1

2(w2 - 2w) = 5 Expand (undo the brackets), and move 5 to left

2w2 - 4w - 5 = 0 a=2, b=-4, c=-5

z(z-1) = 3 Expand, and move 3 to left

z2 - z - 3 = 0 a=1, b=-1, c=-3

5 + 1/x - 1/x2 = 0 Multiply by x2 5x2 + x - 1 = 0 a=5, b=1, c=-1

The solutions to a Quadratic Equation are where it equals 0

There are 3 methods to finding solutions:

1. Factoring the Quadratic (see below);2. Completing the Square (see below); or3. Using the special Quadratic Formula:

Plus/Minus

In the third method listed, with the ±, this means there are literally two answers. Each answer will be a point on the graph. For example:

“b2 – 4tac” is what we call a discriminant because it discriminates between the possible answers:

When b2 – 4ac is positive, you get two real solutions

When it is 0, you get one real solution (both answers are the same) When it is negative you get two complex solutions (these use imaginary numbers, because

the square root of a negative number is an invalid input).

But, look at how we use the quadratic formula to really understand how this all works:

Example: Solve 5x² + 6x + 1 = 0

Co-efficients are: a = 5, b = 6, c = 1

Quadratic formula is: x = [ -b ± √(b2-4ac) ] / 2a

Put in the values: x = [ -6 ± √(62-4×5×1) ] / (2×5)

x = [ -6 ± √(36-20) ]/10

x = [ -6 ± √(16) ]/10 here we see that it is positive, so we will get two real solutions

x = ( -6 ± 4 )/10 this means that we do both possible scenarios: -6 + 4 and =6 – 4, then divide both answers by 10

x = -0.2 or -1 this is our answer that we can see on the graph

We can check that this is correct; up above, I said that the golden rule is ax2 + bx + c = 0

So

Complex Solutions

Above it is said that that if the discriminant (the value of b2 – 4ac) is negative you get complex solutions

Example: Solve 5x2 + 2x + 1 = 0

Coefficients are: a = 5, b = 2, c = 1 The discriminant here is a negative number: b2 - 4ac = 22 - 4×5×1 = -16

Apply the quadratic formula: x = [ -2 ± √(-16) ] / 10

Note that the square root of -16 is an imaginary number, so we get: x = ( -2 ± 4i )/10

Answer: x = -0.2 ± 0.4i

Examples

1. Solve the quadratic equation 6x 2 + 7x - 3 = 0

2. Solve the equation x(10x - 1) = 2

Step 1 (write the base equation): x(10x - 1) = 2

Step 2 (expand the brackets, and move the 2): 10x2 - x - 2 = 0

3. Solve the quadratic equation 15x 2 - 26x - 21 = 0

Note that it is (4 x 15 x -21) NOTE -4 x 15 x -21

4. Solve the quadratic equation 21x 2 - 12x + 1 = 0

= 0.101 or 0.470 correct to 3 decimal places

5. Solve the quadratic equation 8x 2 + 25x + 13 = 0

= -2.466 or -0.659 correct to 3 decimal places

6. Solve the quadratic equation x 2 - 6x + 25 = 0

3 + 4i or 3 - 4i (note that we get this answer because we are dividing both the “6” and the ±8i by 2)

7. Solve the quadratic equation y 2 + 10y + 29 = 0

= -5 + 2i or -5 - 2i

8. Solve the quadratic equation x 2 - 6x + 13 = 0

= 3 - 2i or 3 + 2i

9. Solve the quadratic equation 2x 2 + 6x + 29 = 0

= -1.5 - 3.5i or -1.5 + 3.5i

Factoring Quadratic EquationsWe factor to find the roots.

We have a pretty simple method of factoring:

Remember that this is the golden rule:

1. Find two numbers that will multiply together to give a x c, and add together to give b2. Rewrite the middle with those numbers3. Factor the terms

Example: 2x2 + 7x + 3

Step 1 (find two numbers that will multiply together to give a x c, and add together to give b. 2 x 3 is 6, and b is 7):  6 and 1 do that (6×1=6, and 6+1=7)

Step 2 (rewrite in the middle): 2x2 + 6x + x + 3

Step 3 (factor the first two and last two terms separately): 2x2 + 6x factors into 2x(x+3), and x+3 doesn’t actually change in this case. So 2x(x+3) + (x+3) = (2x+1)(x+3)

Check: (2x+1)(x+3) = 2x2 + 6x + x + 3 = 2x2 + 7x + 3 (Yes)

Example: 6x2 + 5x – 6

Step 1 (find the numbers that multiply together to give a x c, and add together to give b): -4×9 = -36 and -4+9 = 5

Step 2 (rewrite 6x with -4x and 9x): 6x2 - 4x + 9x – 6

Step 3 (factor the first two and last two): 2x(3x - 2) + 3(3x -2) = (2x+3)(3x - 2)

Check: (2x+3)(3x - 2) = 6x2 - 4x + 9x - 6 = 6x2 + 5x - 6 (Yes)

Why Factor? It helps find the roots

Factoring helps us find the roots of the quadratic equation (where the equation is 0. All you have to do (after factoring) is find where each of the two factors (left and right) become 0.

For example, if you had to find the roots of 6x2 + 5x -6. We already know that the factors are (2x + 3)(3x – 2). So we can figure out the roots pretty easily.

(2x + 3) would be 0 when x = -3/2

(3x – 2) would be 0 when x = 2/3

Therefore, the roots of 6x2 + 5x - 6 are: -3/2 and 2/3

Finding the factors using the quadratic formula

Using this formula, you can get the two answers x+ and x− (one is for the "+" case, and the other is for the "−" case in the "±"), and you can get this factoring: a(x - x+)(x - x−)

Examples

1. Factor 6x 2 + x - 2

2. Factor 6x 2 - 47x + 77

3. Factor 6x 2 + 7xy - 20y 2

4. Find the roots of 20x 2 - 13x + 2 = 0 using factoring.

5. Find the roots of 8x 2 + 14x - 15 = 0 using factoring.

6. Solve

7. Factor 8y 2 - 2y – 15

8. Factor 15x 2 + 14x – 16

9. Find the roots of 6x 2 - x - 15 = 0 using factoring.

10. Find the roots of 8y 2 + 10y - 25 = 0 using factoring.

11. Solve (3x + 2)(x - 5) = 28

Completing the SquareThis method can be used for any quadratic equation. It’s actually the basis for the quadratic equation.

Remember the golden rule:

Completing the square allows you to take the “x2 + bx” portion of this formula and reduce it to just one x. Below is a geometrical explanation:

In practice, we add (b/2)2 to our formula, and it will look like this:

x2+ bx + (b/2)2 = (x + b/2)2

But remember, we will need to subtract (b/2)2 as well, to keep the balance. For example:

:

Solving General Quadratic Equations by Completing the Square

In this picture we can see that we have to shapes representing the first two parts of the formula; x2 + bx

We want to max this shape as close to a square as possible. If we left the shape as it was, it’d clearly be a rectangle. But we can almost make a square if we split ‘bx’ in half and add one of these halves to the bottom, as is pictured here.

The blue outline represents our new shape. The green square is what we need to make a complete square. This is what we add into our formula. .

One thing important to note here is that in our golden rule there is an “a”. As you can see in our method above, we are not dealing with any coefficient. So if there is a coefficient in place of ‘a’, you need to divide all terms by ‘a’ (the coefficient of x2). So:

=

There are 5 steps to solve a quadratic equation with this method:

1. Divide all terms by a (the coefficient of x2) to remove “a”.2. Move (c/a) to the right side of the equation, where the 0 is.3. Complete the square on the left hand side of the equation by adding (b/2)2, and balance this

by adding it to the right side of the equation (this will give you the vertex [turning point] of the equation)

4. Take the root of both sides of the equation5. Subtract the number that remains on the left hand side of the equation to isolate x

Examples:

1. x2 + 4x + 1 = 0

Step 1 (we actually go straight to step 2 in the list of 5 steps, as there is no coefficient for x2): x2 + 4x + 1 = 0

Step 2 (move the number term (c/a), in this case just c, to the right side of the equation): x2 + 4x = -1

Step 3 (complete the square by adding (b/2)2 to both sides): x2 + 4x + 4 = -1 + 4. This can be simplified into (x + 2)2 = 3

Step 4 (take the square root of both sides of the equation): x + 2 = ±√3 = ±1.73

Step 5 (subtract 2 from both sides, doing both alternatives for 1.73): x = ±1.73 – 2 = -3.73 or -0.27

As above, remember that the end of the 3rd step gives you the vertex (turning point) of the equation)

2. Solve 5x2 – 4x – 2 = 0

Step 1 (divide all terms by 5): x2 – 0.8x – 0.4 = 0

Step 2 (move c to the right side of the equation): x2 – 0.8x = 0.4

Step 3 (complete the square by adding (b/2)2 to both sides): (b/2)2 = (0.8/2)2 = 0.42 = 0.16

THEREFORE: x2 – 0.8x + 0.16 = 0.4 + 0.16

WHICH CAN BE FACTORED INTO: (x – 0.4)2 = 0.56. You get this factoring because (x – 0.4)2 = (x – 0.4)(x – 0.4) = x2 -0.8 + 16

Step 4 (take the root of both sides of the equation): x – 0.4 = ±√0.56 = ±0.748 (to 3 decimals)

Step 5 (subtract -0.4 from both sides [this means add 0.4]): x = ±0.748 + 0.4 = -0.348 or 1.148

Examples

1. What number should be added to to complete the square?

2. What number should be added to 9y 2 + 1.2y to complete the square?

Step 1 (divide the terms by the coefficient of y2):

Step 2 (calculate (d/2)2): THEREFORE

Step 3 (bring back the coefficient of 9): 9 • 1/225 = 9/225 = 1/25 = 0.04

Therefore 0.04 needs to be added: 9y2 + 1.2y + 0.04 = (3y + 0.2)2

3.

4. In solving the quadratic equation 2x 2 - 12x + 13 = 0 by the method of Completing the Square, which of the following is correct?

Step 1 (remove the coefficient 2 by dividing each term by 2): x2 - 6x + 6.5 = 0

Step 2 (move the 6.5 to the other side): x2 - 6x = -6.5

Step 3 (add (b/2)2): b =-6 SO (d/2)2 = (-3)2 = 9 THEREFORE x2 - 6x + 9 = -6.5 + 9

Step 4 (factor x2 - 6x + 9): (x - 3)2 = 2.5

5. In solving the quadratic equation 5x 2 + 2x - 9 = 0 by the method of Completing the Square, which of the following is correct?

Step 1 (divide all terms by 5): x2 + 0.4x - 1.8 = 0

Step 2 (put -1.8 on the right side): x2 + 0.4x = 1.8

Step 3 (complete the square by adding (b/2)2 to each side): (b/2)2 = (0.2)2 = 0.04 THERFORE x2 + 0.4x + 0.04 = 1.8 + 0.04

Step 4 (factor x2 + 0.4x + 0.04): (x + 0.2)2 = 1.84

6. In solving the quadratic equation 4y 2 - 12y + 7 = 0 by the method of Completing the Square, which of the following is correct?

Step 1 (divide all terms by 4): y2 - 3y + 1.75 = 0

Step 2 (move 1.75 to the other side): y2 - 3y = -1.75

Step 3 (complete the square by adding (b/2)2to both sides): (b/2)2 = (-1.5)2 = 2.25 THEREFORE y2 - 3y + 2.25 = -1.75 + 2.25 = 0.5

Step 4 (factor y2 - 3y + 2.25. I’ve found a handy way of finding what numbers to use when factoring. Divide b by 2, and use that number. Here you’d get -1.5, so you’d get): (y - 1.5)2 = 0.5

7. In solving the quadratic equation 8z 2 - 3z - 2 = 0 by the method of Completing the Square, which of the following is correct?

Step 1 (divide all terms by 8): z2 - 0.375z - 0.25 = 0

Step 2 (move the -0.25 to the right side): z2 - 0.375z = 0.25

Step 3 (complete the square by adding (b/2)2 to each side): (b/2)2 = (-0.1875)2 = 0.03515625 THEREFORE z2 - 0.375z + 0.03515625 = 0.285152625

Step 4 (factor. Remember to divide b by 2, which = -0.1875): (z - 0.1875)2 = 0.28515625

FunctionsWhat is a function? A function relates and input to an output. It relates each element of a set with exactly one element of another set (possibly the same set).

A function is commonly written as ‘f(x) = …’.

Input, Relationship, Output

The three main parts to a function are:

The input The relationship The output

Example:

x2 (squaring) is a function

x3+1 is also a function Sine, Cosine and Tangent are functions used in trigonometry

You would say “f of x equals x squared”: Here we can see that the function is called f and “x” goes in

f(x) = x2 shows you that the function “f” takes “x” and squares it.

So, for the function ‘f(x) = x2’, where x = 4, then the output = 16. In fact we can write f(4) = 16

‘x’ is just a place holder, the function f(x) = 1 - x + x2 is just the same as writing:

f(q) = 1 - q + q2

h(A) = 1 - A + A2

w(θ) = 1 - θ + θ2

Sometimes There is NO Function Name

Some functions have no name, for example y = x2. There is still:

an input (x) a relationship (squaring) an output (y)

Relating

Remember, a function relates an input to an output.

Saying “f(4) = 15” is like saying 4 is somehow related to 16.

What Types of Things Do Functions Possess?

Numbers are obvious. They can contain letters or ID codes too. E.g. ("A"→"B") OR ("A6309"→"Pass")

But more importantly, a function takes elements of a set (a collection of things), and gives back elements of a set. For example, a set of even numbers like {2, 4, 6} can have an output of {6, 12, 18}. If you multiplied each number in the first set by 3, you’d get the numbers in the second set.

Each individual thing in the set (such as “4” or “hat”) is called a member, or element.

A Function is Special

Functions have special rules:

A function must work for every possible input value You can only have one relationship for each input value (an input an only equal 1 number).

HOWEVER two inputs are allowed to equal the same output.

The Two Important Things!

Remember the formal definition of a function is A function relates each element of a setwith exactly one element of another set (possibly the same set).

So we get two important things to note from this:

1. “… each element…” means that every element in x is related to some element in ya. We can say that the function covers X (relates every element of it) BUT some

element of y might not be related to x at all, which is fine.2. “… exactly one…” means that a function is single valued. It will not give back 2 or more

results for the same input.

If these two rules aren’t met, it’s not a function. There’s still a relationship, but it’s not a function.

Vertical Line Test

The idea of a function being single valued means that, if you drew a vertical line anywhere on a graph, the line of the graph would not cross this vertical line twice. Example:

Infinitely Many

Functions usually works on sets with infinitely many elements.

Domain, Codomain and Range/All The Names Used For Elements of a Set

Ordered Pairs

You can also write functions as “ordered pairs”, such as (4, 16). This is very convenient for graphing. (4, 16) means that the function takes in “4” and gives out “16”.

A function can therefore be defined as a set of ordered pairs:

But remember that the function must be single valued. Therefore if it contains (a, b) and (a, c), then b must equal c. This is logical, because a cannot equal both b and c if b and c are different numbers. An input cannot produce two different results. Example: {(2,4), (2,5), (7,3)} is not a function because {2,4} and {2,5} means that 2 could be related to 4 or 5.

Ordered pairs are easy to plot, as they are also coordinates:

A Function Can Be In Pieces

Note that you can create functions that behave differently depending on the input value. This is called a piecewise function: http://www.mathsisfun.com/sets/functions-piecewise.html

Explicit vs. Implicit

An explicit function is one that shows you how to go directly from x to y, such as: y = x3 – 3 (when you know x, you can find y). This is the classic y = f(x) style.

An implicit function is one where it is not given directly, such as x2 – 3xy + y3 = 0 (you may be able to find x, but that won’t help finding y).

Examples

1. The function f is defined on the real numbers by f(x) = 2 + x - x 2 What is the value of f(-3)?

Step 1 (substitute ‘x’ for ‘-3’): 2 + -3 - -32 = 2 -3 – 9

Step 2 (do the arithmetic): -10

2. The function g is defined on the real numbers by g(x) = (x 2 + 1)(3x - 5) What is the value of g(4)?

Step 1 (substitute ‘x’ for ‘4’): (42 + 1)(3 x (4) - 5)

Step 2 (work out the brackets): (17) x (7)

Step 3 (do the math): 119

3. Which one of the following relations is not a function?

Answer: B. In B, the numbers 1 and 2 are both related to more than one number, so this cannot be a function.

4. The function f is defined on the real numbers by f(x) = x 2 - x – 10 If f(a) = -4, what is the value of a?

Step 1 (write the base equation): x2 - x – 10 = -4

Step 2 (move the -4 into the equation): x2 - x – 10 + 4 = 0

Step 3 (add the like terms. Note that -10 + 4 = -6): x2 - x – 6

Step 4 (factor the equation): (x – 2)(x + 3)

Therefore, a = -2 OR 3

5. The function f is defined on the real numbers by f(x) = 2x 2 - 5x + 12 If f(k) = 10, what is the value of k?

Step 1 (write the base equation): 2x2 - 5x + 12 = 10

Step 2 (move the ‘10’ into the equation): 2x2 - 5x + 12 – 10 = 0

Step 3 (add the like terms): 2x2 - 5x + 2 = 0

Step 4 (factor the equation): (2k - 1)(k - 2)

6.

To see if this is true, take on element of the set and put it into the function. For example, -22 – 9 = -5, therefore we can see the function is f(x) = x2 – 9

7.

SequencesA sequences is a set of things that are in order. There is a relationship/pattern between the numbers.

There is a relationship of “2” here, that is, each number increases by 2 each time. You could write this sequence as:

Xn = 2n + 1. To see why this would work, we let “n” equal the term we wish to find. We multiply this by the “common difference”/relationship, which is 2, and we add 1. So, take the third term for example, we would write this as ‘2 x 3 + 1 = 7’. Look at our sequence above, and we can see this is true.

Examples of sequences:

{1, 2, 3, 4 ,...} is a very simple sequence (and it is an infinite sequence) {20, 25, 30, 35, ...} is also an infinite sequence {1, 3, 5, 7} is the sequence of the first 4 odd numbers (and is a finite sequence) {4, 3, 2, 1} is 4 to 1 backwards {1, 2, 4, 8, 16, 32, ...} is an infinite sequence where every term doubles {a, b, c, d, e} is the sequence of the first 5 letters alphabetically {f, r, e, d} is the sequence of letters in the name "fred" {0, 1, 0, 1, 0, 1, ...} is the sequence of alternating 0s and 1s (yes they are in order, it is an

alternating order in this case)

So we can see that, though they are in order, you can say what order that will be. They could go forwards, backwards, alternating, and so on.

Like a set

Sequences are like sets (above), EXCEPT:

The terms are in order (sets don’t have to have ordered terms) The same term can appear multiple times in a sequence, but only once in sets.

o Example: {0, 1, 0, 1, 0, 1, ...} is the sequence of alternating 0s and 1s. The set would be just {0,1}

Notation

A sequence uses the same notation as sets: each element is listed, separated by a comma, then curly brackets (also called set brackets or braces) are put around the entire sequence/set: {3, 5, 7, …}

A Rule

A sequence usually has a rule, which helps to find the value of each term, for example, in the above example, our rule was Xn = 2n + 1, which contains our rule.

As a Formula

We’ve already seen that we can use formulas to find terms, here’s a further explanation.

“n” is the term you are trying to find.

A rule for {3, 5, 7, 9, ...} can be written like this xn = 2n + 1. We can see that the difference between these numbers is “2”. If we want to find the 10th term in the sequence, we multiply 10 by 2 (the difference) and add 1 = 21.

Another example. Calculate the first 4 terms of this sequence: {an} = { (-1/n)n }

Arithmetic Sequences (AKA Arithmetic Progressions)In an arithmetic sequence, the difference between one term and the next is constant. What’s interesting is that there are many ways to write a rule to get the same result.

Take the following sequence for example: 1, 5, 7, 10, 13, 16, 19, 22, 25, …

The sequence has a difference of 3 between each number. You could write its rule as xn = 3n – 2

In general, you could write the sequence like this: {a, a+d, a+2d, a+3d, ... }

Where a is the first term of the sequence, and d is the difference between the terms (called the common difference)

Though you can write the sequence with the above method, that’s not our rule. You can make this rule by xn = a + d(n – 1) : (We use "n-1" because d is not used in the 1st term).

Summing an Arithmetic Series

Remember one way we can write an arithmetic series (this is not the rule) is by a + (a+d) + (a+2d) + (a+3d) + ...

We can sum up this formula (add up the value of all the terms) with the following formula:

What this means:

Example: add up the first 10 terms of the arithmetic sequence {1, 4, 7, 10, 13, …}

The values of a, d and n are:

a = 1 (the first term) d = 3 (the "common difference" between terms) n = 10 (how many terms to add up)

Our base formula:

∑k=0

n−1

(a+kd )=n2

(2a+(n−1 )d )

Becomes:

∑k=0

10−1

(a+k ∙3 )=102

(2∙1+(10−1 ) ∙ d )

Which is 5(2+9·3) = 5(29) = 145

Why does this formula work?

Examples

1. What is the fiftieth term of the arithmetic sequence 3, 7, 11, 15, ... ?

Step 1 (write the rule in the form of xn = a + d(n-1)): xn = 3 + 4(50 – 1)

Step 2 (do the arithmetic, working out the brackets first): 199

2. What is the twentieth term of the arithmetic sequence 21, 18, 15, 12, ... ?

Step 1 (write the rule in the form of xn = a + d(n-1) but note that the common difference, d, is -3, as we are descending in order): 21 + -3(20 - 1)

Step 2 (do the arithmetic, working out the brackets first): -36

3. What is the sum of the first sixteen terms of the arithmetic sequence 1, 5, 9, 13, ... ?

Step 1 (write the summation formula): Sn = n/2(2a + (n - 1)d)

Step 2 (substitute the terms): S16 = 16/2(2 × 1 + 15 × 4)) = 8(2 + 60) = 8 × 62 = 496

4. What is the sum of the first thirty terms of the arithmetic sequence 50, 45, 40, 35, ... ?

Step 1 (write the summation formula): Sn = n/2(2a + (n - 1)d)

Step 2 (substitute the terms and do the arithmetic. Note that d here is -5 as we are descending in order): 30/2(2 × 50 + 29 × -5)) =15(100 - 145) = 15 × -45 = -675

4. What is the sum of the eleventh to twentieth terms (inclusive)of the arithmetic sequence 7, 12, 17, 22, ... ?

Step 1 (find the sum of the first 10 numbers):

Step 2 (find the sum of the first 20 numbers):

Step 3 (subtract the sum of the first 10 numbers from the sum of the first 20 numbers):

5. The fifth term of an arithmetic sequence is 11 and the tenth term is 41. What is the first term?

Step 1 (we first need to work out what the difference between each term is, we know the 10th term is 41 and the fifth is 11, find their difference and divide by how many number separate them in the sequence): (41-11)/5 = 6. So there is a difference of 6

The mathopolis way of finishing this is:

To get this answer, all I did was 11 + (-6 x 4) = -13

6. The eleventh term of an arithmetic sequence is 30 and the sum of the first eleven terms is 55. What is the common difference?

Step 1 (use the formula for the n’th term for the value of the 11th term. We already know it’s 30, but you need to write it out, you’ll see why): x11 = a + d(11 - 1) = 30 WHICH BECOMES a + 10d = 30

Step 2 (now find the sum of the first 11 terms. We already know it’s 55, but we still need to find it, you’ll see why soon):

7. How many terms of the arithmetic sequence 2, 8, 14, 20, ... are required to give a sum of 660?

To see why we remove the 2 from the denominator of 2 and apply it to the terms in brackets, pretend n = 5. Both methods would give the answer of 70, it’s just easier to remove the 2 and get rid of the fraction and work with smaller numbers.

Geometric SequencesGeometric sequences are sometimes called Geometric Progressions

In geometric sequences each term is found by multiplying the previous term by a constant.

For example, in the sequence 2, 4, 8, 16, 32, 64, 128, 256, … you could write the rule as xn = 2n as the sequence has a factor of 2 between each number

In general you could write a geometric sequence like this: {a, ar, ar2, ar3, …} where a is the first term, and r is the factor between the terms called “the common ratio”.

The rule for this would be xn = ar(n – 1)

Note that r shouldn’t be -. When r = 0, you get the sequence {a, 0, 0, …} which is not geometric

Geometric sequences can also have progressively smaller numbers:

We call them geometric sequences because they work like increasing dimensions in geometry:

Summing a Geometric Sequence

To sum: a + ar + ar2 + ... + ar(n-1) – each term is ark, where k starts at 0 and goes up to n-1

This works because we are changing the value of the exponent progressively. We would therefore start off with ar0 which is the same as just writing a. Then we get ar1, which would just be the same as ar. Then, from the third number in the sequence onwards, we would start using ar(n-1)

Use this formula to sum:

Why Does This Work?

First, we will call the whole sum “S”. So S (every term) = a + ar + ar2 + ... + ar(n-2)+ ar(n-1)

Next we will multiply s (every term) by r, giving us ar + ar2 + ar3 + ... + ar(n-1) + arn

Third, subtract them:

Therefore, S - S·r = a - arn

So we can rearrange to find/isolate “S”. Factor out S and a:

S(1- r) = a(1- rn)

Now, divide by (1 – r): S = a(1- rn)/(1- r)

This is our formula:

Infinite Geometric Sequences

BUT NOTE: this does not work where r is more than 1 or less than -1

Recurring Decimal

Examples

1. What is the eleventh term of the geometric sequence 3, 6, 12, 24, ... ?

Step 1 (figure out the correct rule. We can see that each term doubles, so the common factor is two. We will need an exponent. The relevant exponent here is ‘n – 1’): xn = arn-1

Step 2 (substitute the relevant numbers into the rule): x11 = 3 × 210 = 3 × 1,024 = 3,072

2. What is the ninth term of the geometric sequence 81, 27, 9, 3, ... ?

Step 1 (figure out the correct rule. Here, the numbers are going down in thirds, so the common ration is ‘1/3’. The exponent will be n-1): xn = arn-1

Step 2 (substitute the relevant numbers into the rule): x9 = 81 × (1/3)8 = 81 × 1/6,561 = 81/6,561 = 1/81

3. What is the sum of the first eight terms of the geometric sequence 5, 15, 45, ... ?

Step 1 (figure out the values of a, r, and n): a = 5 (the first term) r = 3 (the "common ratio") n = 8 (we want to sum the first 8 terms)

Step 2: NOTE this formula should be a[(1 - rn)/(1-r)]

Step 3 (put the relevant numbers into the formula):

4. What is the sum of the first nine terms of the geometric sequence 20, 10, 5, ... ? Give your answer as a decimal correct to 1 decimal place.

Step 1 (figure out the values of a, r, and n): a =20 (the first term) r = 1/2 (the "common ratio") n = 9 (we want to sum the first 9 terms)

Step 2:

5. The fourth term of a geometric sequence is 27 and the seventh term is 1. What is the first term?

6. The first term of a geometric sequence is 192 and the fifth term is 0.75. What is the common ratio?

7. Add up all the terms of the following infinite geometric sequence:

8. How many terms of the geometric sequence 2, 8, 32, 128,... are required to give a sum of 174,762 ?

Step 1 (determine the values of a and r): a = 2 (the first term) r = 4 (the "common ratio") We also know the sum to n terms Sn = 174,762

Step 2 (Substitute these numbers into the formula):

Step 3 (start beginning to isolate the ‘4n’. Start first by removing the denominator by multiplying both sides by 3):2( 4n – 1) = 174,762 × 3 = 524,286

Step 4 (remove the 2 by dividng both sides by 2): 4n – 1 = 262,143

Step 5 (remove the ‘-1’ by adding 1 to both sides): 4n = 262,144

Step 6 (now we have to use a logarithm to figure out what our exponent is equal to): n = log4(262,144) = ln(262,144)/ln(4) = 12.476.../1.386... = 9

8. The sum of the first two terms of a Geometric Progression is 36 and the product of the first and third terms is 9 times the second term. Find the sum of the first 8 terms.

Fibonacci Sequences

Triangular Sequences

Sequences – Finding a RuleTo predict a missing number/numbers in a sequence, you need to have a rule.

For example, what are the next 3 numbers in the sequence 1, 4, 9, 16? Our answer is that they are squares of numbers starting from 1 and moving upwards (12=1, 22=4, 32=9, 42=16, ...). Therefore our sequence is 1, 4, 9, 16, 25, 36, 49, ... And we can write the rule as xn = n2

Note xn means "term number n", so term 3 would be written x3. x3 = 32 = 9

Another example: 3, 5, 8, 13, 21…?

Our rule is xn = xn-1 + xn-2. So, if we want to calculate the 6th term in this sequence, we write: x6 = x6 – 1 + x6 – 2 = 5 (the fifth term in our sequence) + 4 (the fourth term in our sequence) = 21 + 13 = 34. As such, we can see that this is a Fibonacci Sequence.

The full sequence is 3, 5, 8, 13, 21, 34, 55, 89, ...

Therefore, we can see that xn – 1 means the previous term behind the one you are wanting to find.

Many Rules

Because of the complexity of mathematics, there can be many differing, yet correct rules, for a sequence. As a general rule of thumb/general rule, choose the simplest rule that makes sense, but also mention that there are other solutions

For example: What is the next number in the sequence 1, 2, 4, 7, ?

1. First Solution

We could apply the rule xn = n (n−1 )2

+1

Applying this to the fifth term, we would get 5 (5−1 )2

+1=11

Applying this to the sixth term, we would get 6 (6−1 )2

+1=16

So, we could repeat this and get a sequence of 1, 2, 4, 7, 11, 16, 22, ...

2. Second Solution

We could apply the rule xn = xn-1 + xn-2 + 1

Applying this to the fifth term, we would get: x5 = x5 – 1 + x5 – 2 + 1 = 4th term + 3rd term + 1 = 7 + 4 + 1 = 12Repeating this, we could find a sequence of 1, 2, 4, 7, 12, 20, 33, …

3. Third Solution

We could apply the rule xn = xn-1 + xn-2 + xn-3

Applying this to the fifth term, we would get x5 = x5-1 + x5-2 + x5-3 = 4th term + 3rd term + 2nd term = 7 + 4 + 2 = 13

Repeating this, we could find a sequence of 1, 2, 4, 7, 13, 24, 44, …

Finding Differences

If you can find the differences between the numbers in a sequence, you can find the rule

Take the example of

We can see that the difference between each number in the sequence is 2n. But 2n is not our rule. Our rule will be written was xn = 2n + 5. To see why, look at the table below:

n: (number in the sequence

1 2 3 4 5

Term 7 9 11 13 152n (2 x position

in sequence)2 x 1 = 2 2 x 2 = 4 2 x 3 = 6 2 x 4 = 8 2 x 5 = 10

Wrong by 5 5 5 5 5So we add 5 to the product of 2 and the position of the number we are trying to find in the sequence.

Second Differences

The differences between the differences of numbers in a sequence can help us find a rule. For example:

The second differences are 1. With second differences you multiply by “n2/2”.

Mathisfun gives a trial and error approach with the tables:

Try n2/2

Here, we are increasing our difference by 0.5 each time, which isn’t helpful.

Now trying n2 / 2 - n/2

We are wrong by 1, so try adding 1 to the formula.

Now, we add one and we get n2 / 2 - n/2 + 1. Note that this can be simplified into n(n-1)/2 + 1

Rule: xn = n(n-1)/2 + 1

Sequence: 1, 2, 4, 7, 11, 16, 22, 29, 37, ...

Examples

1. What is the next number of the sequence 1, 3, 7, .... , given that the rule for the sequence is:x1 = 1 AND xn = xn-1 + 2n-1 , n ≥ 2

Step 1 (write the rule): xn = xn-1 + 2n-1

Step 2 (check that the rule works, do this for the 2nd and 3rd terms):

x2 = x1 + 21 = 1 + 2 = 3 is correctx3 = x2 + 22 = 3 + 4 = 7 is correct

Step 3 (work out the 4th term): x4 = x3 + 23 = 7 + 8 = 15

2. What is the next number of the sequence 0, 2, 5, 9,... , given that the rule for the sequence is: x n = ½(n 2 + n - 2)

Step 1 (write the rule): xn = ½(n2 + n - 2)

Step 2 (check it works for the terms we know):

x1 = ½(12 + 1 - 2) = ½(1 + 1 - 2) = 0 is correctx2 = ½(22 + 2 - 2) = ½(4 + 2 - 2) = 2 is correctx3 = ½(32 + 3 - 2) = ½(9 + 3 - 2) = 5 is correctx4 = ½(42 + 4 - 2) = ½(16 + 4 - 2) = 9 is correct

Step 3 (work out the 5th term): ½(52 + 5 - 2) = ½(25 + 5 - 2) = 14

3. Use differences to find the rule for the sequence {0, 2, 6, 12, 20, ...}

4. Use differences to find the rule for the sequence {1, 5, 14, 28, 47, ...}

5. Use differences to find the rule for the sequence {1, 5, 12, 22, 35, ...}

Note that we get the “-n/2” part of the equation because, in our first attempt, we are always wrong by exactly half of n. So divide it by 2.

6. Use differences to find the rule for the sequence {3, 11, 24, 42, 65, ...}

7. Use differences to find the rule for the sequence {3, 13, 30, 54, 85, ...}

8. Use differences to find the rule for the sequence {0, 2, 8, 20, 40, 70, ...}.

9. Find a rule for the sequence {1, 4, 10, 20, 35, 56, ...} using first, second and third differences

10. Use differences to find the rule for the sequence {2, 7, 16, 30, 50, 77, ...}

Other Sources

Simplifying Algebraic Equations

Combining Like TermsTerms: terms are when we have variables and/or numbers multiplied together For example we might see 5a2b numbers and variables multiplied together. “6” is a term too

Like terms: the variables and exponents match.

When we have like terms we can add the coefficients (the number in front) of the like terms

E.g. if you have 5x – 3x. Think of this as saying we have 5 of something and we are taking away 2 more of that something, we get 2 of that something.

For example, if we had 2x + 3x, that’s the same as say (x + x) + (x + x + x), we could just write this as 5x.

Another example, if we had 7y + 2x + 3x + 2y, we can’t add the y’s and the x’s because they may represent a different number. What we can do is add the y’s together and the x’s together, which gives us 9y + 5x.

Examples

1. 4x 3 -2x 2 +5x 3 +2x-4x 2 -6x

We can see that “4x3” and “5x3” are like terms because the variables (the letters) and the exponents (the powers) match. So we can add the numbers in front and we get 4x3 + 5x3

= 9x3.

Next, there’s a “-2x2” (notice the negative stays with the number). The like term is -4x2. Adding the like terms, we get -2x2 + -4x2 = -6x2.

We also have 2x and 6x. It is +2x - -6x = -4x

So the formula is simplified to 9x3 – 6x2 – 4x

2. 4y – 2x + 5 – 6y + 7x – 9

Start with the “4y”. The term like it is -6y. 4y – 6y = -2y.

-2x can be combined with 7x. -2 + 7 = 5x

5 has no variables, just like -9: 5 – 9 = -4.

Solution: -2y + 5x – 4.

3. –m + 3 – 6m

Step 1 (write the base equation): –m + 3 – 6m

Step 2 (put the two ‘m’ terms together): -7m + 3 that is our answer

Distributive propertyDistributive property: if we have a number in front of parenthesis, we take the front number and multiply it by what’s in the parentheses. E.g. a(b + c) = ab + ac.

We use the distribute property to clear parentheses.

For example, saying 2(3x + 5) is the same as saying 2 lots of 3x + 5. This simplifies into 6x + 10.

Or, if we had an issue like 7(3y – 5) – 2(10 + 4y).

Do the left hand side first. Distribute the 7 to the brackets, which will give us 21y – 35.

Now let’s do the right hand side first. Note that we are distributing a negative 2 = -20 – 8y.

We’re not done yet though, we have like terms. We’ve got 21y – 35 – 20 -8y. So we get 13y - 55

Examples

1. -2(5x – 4y + 3)

Take the negative 2 and multiply it by each term in the parenthesis.

Start with -2 x 5x = -10x

Next, multiply the negative 2 by the next term (-4y) = -2 x -4y = 8y

-2 x 3 = -6

Solution: -10x + 8y – 6

2. 4x(7x 2 – 6x + 1)

Start by doing 4x multiplied by 7x2 = 28x3 (we get a 3 because there were already 2 x’s on 7x, and we added another one).

4x x -6x = -24x2

4x x 1 = 4x

Solution: 28x3 – 24x2 + 4x

3. 3 + 7 (-6m – 8)

Step 1 (write the base equation): 3 + 7 (-6m – 8)

Step 2 (distribute the 7 do not distribute ‘10’): 3−42m−56

Step 3 (add the 3 to the like term, which is -56): -42m – 53

4. n + 4(2n – 2)

Step 1 (write the base equation): n + 4(2n -2)

Step 2 (distribute the ‘4’): n + 8n – 8

Step 3 (add the ‘n’ to the like term, which is ‘8n’): 9n – 8

5. 5 (− 4 − 2 m )+ 3 ( m + 7 )

Step 1 (write the base equation): 5(−4−2m)+3(m+7)

Step 2 (distribute the right hand side first): −20−10m+3(m+7)

Ste p 3 (distribute the left hand side): 20−10m+3m+21

Step 4 (re-write the expression so the m terms are together): −10m+3m−20+21

Step 5 (combine): −7m+1

6. 16 – (-v – 8)

Step 1 (write the base equation): 16 – (-v – 8)

Step 2 (the minus sign in front of the parentheses means we must multiply the terms inside by ‘-1’): 16−1(−v−8) = 16+v+8

Step 3 (rewrite the expression to group the numeric terms): v+16+8 = v + 24 and that is our answer

Evaluating expressionsA variable is a letter that represents an unknown number.

A dozen is the same as 12. You could say I have a dozen donuts. Or we can replace word “dozen” with the number it represents, so you’d say “I have 12 donuts”. Similarly, to evaluate algebraic expressions, we can replace the variable with the number it represents. Once we do this, it becomes an order of operations problem. We need to put the number in parentheses.

To evaluate: replace the variable with the number it represents.

Example

1. 4x 2 – 3x + 2 when x = -3

Rewrite, replacing the “x” with “-3” = 4(-3)2 – 3(-3) + 2. Now it becomes an order of operations problem

First, we know to do exponents. 4(9) – 3(-3) + 2. Next we multiply and divide from left to right

36 – 3(-3) + 2. Continuing to multiply, we get to -3 x -3 = +9, so we now get:

36 + 9 + 2 = 47

2. 4b(2x + 3y) when b = -2, x = 5, y = -7

4 (-2) [2(5) + 3 (-7)]. Order of operations tells us to work inside the parentheses first, multiplying from left to right. So we now get:

4 (-2) [10 + 3 (-7)]

4 (-2) [10 – 21]. Now finish the parentheses

4(-2) [-11]. Finish by multiplying from left to right

-8 x -11 = 88.

Writing expressions1. Write an expression to represent: Five more than twice a number x .

We know that twice x can be written as 2x. Now we add 5 to it, which is 5 + 2x

2. Write an expression to represent: Six less than three times a number x .

Three times a number x can be written as 3x. Six less than something means that we subtract 6 from it. If we subtract 6 from 3x, we have 3x−6.

3. The product of three and a number x .

3x

4. Write an expression to represent: The sum of nine and twice a number x .

2x + 9

5. Write an expression to represent: One minus four times a number x .

1 – 4x

6. Write the following: The sum of 1 and the product of 5 and x . Now consider the following: − 8 plus the quantity of − 6 times that expression

−6(5x+1)−8 – note that we get ‘-8’ because adding a negative is just the same as subtracting

7. Take the product of − 5 and x and add − 1 . Now consider: The sum of 3 and the product of − 7 and that expression.

−7(−5x−1)+3

Variables, Expressions and EquationsVariables are just values in expressions that can change. For example “x + 5” is an expression, and can take on a value depending on what ‘x’ is. If ‘x’ was ‘1’, the expression would be equal to 6.

An expression is just a statement of a value. For example x + 5. The value of this expression depends on the variable.

An equation is where you equate two expressions (that is, you are saying one expression is equal to another expression). For example x + 3 = 1. Another example is x + y + z = 5.

A variable can take on a different value depending on the context of the situation. If we had the expression xy and x = 5, y =2. Our expression now becomes 52 = 25. But we could change both of these values if we wanted to.

Simple equations7x = 14. This is the exact same as saying ‘7 times x will equal 14’. To work this out, we would need to divide both sides by 7. The left hand side will just become ‘x’ and the right hand side will become ‘2’. Therefore, x = 2.

Another example ‘3x = 15’. You could divide both sides by 3, or you could multiply both sides by 1/3. Because 1/3rd of 3 = 1, and 1/3rd of 15 = 5.

2y + 4y = 18. First, add the 2y and the 4y together, which is 6y. Remove the 6 by dividing each side by 6. We get our answer, y = 3.

Misc/Jumbled Algebra Problems and Solutions

For this one see here for why

Polynomials

Word Problems1. Imran is 18. Diya is 2. How many years will it take for Imran to be 3 times as old as Diya?

We need to write this as an equation first. Let ‘y’ be the length of time it would take in years for Imran to be 3 times as old as Diya. This time in years will be added onto Imran’s and Diya’s current age, and Imran has to be 3x Diya’s age after this time passes, so our equation is:

18 + y = 3 (2 + y)

18 + y = 6 + 3y

12 + y = 3y

12 = 2y

6 = y

Therefore it will take Imran 6 years to be 3x Diya’s age.

2. In 40 years, Imran will be 11 times as old as he is right now. How old is he right now?

Let ‘x’ be Imran’s current age.

X + 40 = 11x

40 = 10x

4 = x

Therefore, Imran is 4

Sourceswww.mathisfun.com

http://www.mathopenref.com/radical-sign.html