Writing Mathematics With Style An (algebraic) expression is any combination of numbers, variables, exponents, mathematical symbols, and mathematical operations, but never include equal signs or inequality signs! An equation is a mathematical statement that two quantities are equal. Therefore, an equation consists of two expressions separated by an equal sign. An inequality is a mathematical statement expressing an order relationship. Therefore, an inequality consists of two (or more) expressions separated by one of the inequality signs. If more than one inequality sign are used, they must be in the same direction! Classify the following as an expression, equation, inequality, or nonsense:

1) 2x 5

2) 2x 5 = 1

3) y – z ≤ 9

4) y – z + 9

5) 1x =

2y

6) 1x ≥

2y

7) 2 ≤ x ≥ 3

8) 2 ≤ x + 3

9) | 2x – 3 | > 1

10) a < b < c

11) 2 – 3(4 + x)2 – 3x(x – 3)

12) 2 – 3(4 + x)2 = 3x(x – 3)

13) x2 = x + 6

14) x2 ≥ x + 6

15) x2 ≤ x + 6

16) x2 ≤ x ≤ 6

17) x2 ≥ x ≤ 6

Simplifying Expressions: You usually simplify expressions by performing a series of mathematical operations to obtain another expression. To express the fact that the new expression is equal to the old expression, use the equal sign (=). Always arrange problems vertically, if possible. In the examples below, start with the given expression, and simplify to the answer. Example 1: Simplify: GivenExpression GivenExpression = expression1 = expression2 = expression3

= answer

GivenExpression = expression1 = expression2 = expression3

= answer

Example 2: Simplify 16 + 8 ÷ 4 • 2 + 4

16 + 8 ÷ 4 • 2 + 4 = 16 + 2 • 2 + 4 = 16 + 4 + 4 = 20 + 4

= 24 Example 3: Simplify (x – y)(x + y – 2)

(x – y)(x + y – 2) = x2 + xy – 2x – xy – y2 + 2y

= x2 – 2x – y2 + 2y

DO NOT:

1. Use bars under equations or expressions to show addition or subtraction.

2. Place extraneous calculations or comments in the body of a simplification or a solution.

Solving Equations and Inequalities: You usually solve equations and inequalities by performing a series of mathematical operation to obtain another equation or inequality. To express the fact that the new equation is implied by the old equation, sometimes we use the implication sign (�), although it is acceptable if problems are arranged neatly. If possible, arrange problems vertically. In the examples below, start with the given equation, and solve. Example 1: Simplify: GivenEquation Either form given below is acceptable. GivenEquation equation1 equation2 equation3

answer

GivenEquation � equation1 � equation2 � equation3

� answer

Example 2: Solve 2x + 1 = 3x – 5

2x + 1 = 3x – 5 2x + 1 – 3x – 1 = 3x – 5 – 3x – 1 –x = –6

x = 6 Example 3: Solve 2x + 1 ≤ 3x – 5

2x + 1 ≤ 3x – 5 2x + 1 – 3x – 1 ≤ 3x – 5 – 3x – 1 –x ≤ –6

x ≥ 6

Types of Numbers Natural Numbers (Counting Numbers) N N = {1, 2, 3, 4, 5, ...} Whole Numbers W W = {0, 1, 2, 3, 4, 5, ...} Integers Z Z = {..., –4, –3, –2, –1, 0, 1, 2, 3, 4, ...} Rational Numbers Q

Q = { ab | a, b, � Z , b ≠ 0}

Irrational Numbers I Numbers that can be written as an infinite nonrepeating decimal Real Numbers R Any number that is rational or irrational (R = Q ≈ I)

Real Number Line Visualize a line with equally spaced markers each of which is associated with the integers. If the integers have their natural order, then the real numbers can be visualized as points on the line.

0 1 2 3-1-2-3 We notice that 1. Every real number corresponds to a unique point on the line. 2. Every point on the line corresponds to a unique real number. This is why the set of real numbers is sometimes referred to as the real number line.

Absolute Value

| x | = x if x ≥ 0–x if x < 0

The absolute value of a real number is equal to its (positive) distance to the origin!

Rule for Order of Operations

(Please Excuse My Dear Aunt Sally)

1. Parentheses: Simplify all groupings first.

2. Exponents: Calculate exponential powers and radicals.

3. Multiplication and Division: Perform all multiplications and divisions as they occur from left to right.

4. Addition and Subtraction: Perform all additions and subtractions as they occur from left to right.

Evaluate the following expressions and note the use of the equal signs because we use mathematics writing style: Ex 1: (4 – 6)2 + 6(–4) + 5

(4 – 6)2 + 6(–4) + 5

(P) = (–2)2 + 6(–4) + 5

(E) = 4 + 6(–4) + 5

(M) = 4 – 24 + 5

(S) = –20 + 5

(A) = –15

Ex 2: 6 + 24 ÷ 3 • 2 + 3 16

6 + 24 ÷ 3 • 2 + 3 16

(E) = 6 + 24 ÷ 3 • 2 + 3 • 4

(D) = 6 + 8 • 2 + 3 • 4

(M) = 6 + 16 + 3 • 4

(M) = 6 + 16 + 12

(A) = 22 + 12

(A) = 34

Ex 3: 2 [ 5 + 2 ( 6 + 3 – 4) ]

2 [ 5 + 2 ( 6 + 3 – 4) ]

= 2 [ 5 + 2 ( 9 – 4) ]

= 2 [ 5 + 2 ( 5) ]

= 2 [ 5 + 10 ]

= 2 [ 15 ]

= 30

Ex 4: 10 + 12 ÷ 4 + 2 • 3

10 + 12 ÷ 4 + 2 • 3

= 10 + 3 + 2 • 3

= 10 + 3 + 6

= 13 + 6

= 19

Types of Intervals

Interval Algebraic Interval Notation Graph Notation Description (a, b)

a b a < x < b Open, finite

[a, b]

a b a ≤ x ≤ b Closed, finite

[a, b)

a b a ≤ x < b Half–open, finite

(a, b]

a b a < x ≤ b Half–open, finite

(a, ∞)

a x > a Open, infinite

(–∞, b)

b x < b Open, infinite

[a, ∞)

a x ≥ a Closed, infinite

(–∞, b]

b x ≤ b Closed, infinite

If your answer is composed of two (or more) distinct intervals, then the algebraic form of your answer must contains the conjunction 'OR'.

Example 1: Describe algebraically the following intervals: a)

-3 4 –3 < x < 4

b)

-1

x ≤ –1

c)

2 8 2 ≤ x < 8

d)

5 x > 5

Example 2: Describe the region(s) containing the indicated sign(s): Sign Graph Answer

+ 0 0+ +–

-4 -1 x < –4 or x > –1

+, 0 0 0+– + +–

2 4 5

(*)

2 < x ≤ 4 or x ≥ 5

–, 0 0 0+– + +–

2 4 5

(*)

x < 2 or 4 ≤ x ≤ 5

– 0 0+ –+– (*)

-2 0 2 –2 < x < 0 or x > 2

Addition Property of Equality For any real numbers a, b, and c,

if a = b, then a + c = b + c. This means that you can always add the same quantity to both sides of an equation. Because of the relation between addition and subtraction, you also can always subtract the same quantity to both sides of an equation. Examples: Solve the following:

x – 4 = 9 x – 4 + 4 = 9 + 4

x = 13

w + 2 = –3 w + 2 – 2 = –3 – 2

w = –5

Multiplication Property of Equality For any real numbers a, b, and c, with c ≠ 0,

if a = b, then a • c = b • c. This means that you can always multiply the same nonzero quantity to both sides of an equation. Because of the relation between multiplication and division, you also can always divide the same nonzero quantity to both sides of an equation. Examples: Solve the following:

x3 = –5

3 • x3 = 3 • (–5) x = –15

4w = 8 4w4 =

84

w = 2

Addition Property of Equations You can always add or subtract the same quantity to both sides of an equation without affecting the solution.

Example 1: Solve 3x – 4 = 2x + 3

We add the quantity (–2x + 4) to both sides of the equation :

3x – 4 = 2x + 3

3x – 4 – 2x + 4 = 2x + 3 – 2x + 4

x =

Multiplication Property of Equations

You can always multiply or divide the same nonzero quantity to both sides of an equation without affecting the solution.

Example 2: Solve –4x = 12

We divide both sides of the equation by –4.

–4x = 12

–4x–4 =

12–4

x = –3

Solving First Degree Equations in One Variable (Be GLAD)

Remember: You can always simplify any side of the equation at any time. 1. Simplify the expressions on both sides of the equation. You must

first eliminate all of the Grouping symbols and simplify if necessary.

2. If there are fractions, you may multiply both sides of the equation

by the LCD. Simplify if necessary. 3. You want the variable terms on one side of the equation and the

constant terms on the other side. If this is not the case, you

should Add or subtract to both sides of the equation a variable term and/or a numerical value in order to get the variable terns on one side and the constant values on the other side. Simplify if necessary.

4. You want the variable term coefficient to be 1. If this is not the

case, you should Divide both sides of the equation by the variable term coefficient. Simplify if necessary.

5. Check your answer by substituting it into the original equation. Note: Always make sure your final answer has the variable on the left side of the equation!!!

Examples: Solve the following and use mathematics writing style:

Ex 1: 3x + 3 = x – 5

3x + 3 – x – 3 = x – 5 – x – 3

2x = –8

2x2 =

–82

x = –4

Ex 3: 3(5y – 2) – 4y = –2(y + 3)

15y – 6 – 4y = –2y – 6

11y – 6 = –2y – 6

11y –6 + 2y +6 = –2y –6 + 2y +6

13y = 0

13y13 =

013

y = 0

Ex 2: x – 5 = 3x – 8

x – 5 – 3x + 5 = 3x – 8 – 3x + 5

–2x = –3

–2x–2 =

–3–2

x = 3/2

Ex 4: 3z4 +

12 = 3 +

z3

12

3z

4 + 12 = 12

3 + z3

9z + 6 = 36 + 4z

9z + 6 – 4z –6 = 36 + 4z – 4z –6

5z = 30

5zz =

305

z = 6

Equations Involving Fractions When we solve equations with fractions, we always assume that no denominator is zero. Therefore, we can find the LCD and multiply both sides by this nonzero factor. We must check to see that our solution does not cause any denominator to be zero.

To solve equations with (simple) fractions: Step 1: Identify all fractions in the equation and find the LCD. Step 2: Multiply the both sides of the equation by the LCD. Step 3: Solve the resulting equation. Step 4: Check the answer into the original problem. (At least

check to make sure no denominator can be zero.)

W A R N I N G : You must know the difference between an expression and an equation. When you solve an equation, you may multiply both sides by the LCD and we get an equation without fractions. When you have an expression with fractions, you must perform the indicated operation and / or simplify. Nonsense like treating expressions like equations will not be tolerated!!

Examples: Solve the following:

a) 2

x – 6 = 3

x – 8

(x – 6)(x – 8) 2

x – 6 = (x – 6)(x – 8) 3

x – 8

2(x – 8) = 3(x – 6)

2x – 16 = 3x – 18

–x = –2

x = 2

b) z – 4

z2 – 2z = 1z –

2z2 –

4z3 – 2z2

z – 4

z(z – 2) = 1z –

2z2 –

4z2(z – 2)

z2(z – 2) z – 4

z(z – 2) = z2(z – 2)

1

z – 2z2 –

4z2(z – 2)

z(z – 4) = z(z – 2) – 2(z – 2) – 4

z2 – 4z = z2 – 2z – 2z + 4 – 4

z2 – 4z = z2 – 4z

0 = 0

all real numbers except 0, 2

c) y – 2y – 3 = 1 –

2y2 – 9

y – 2y – 3 = 1 –

2(y + 3)(y – 3)

(y + 3)(y – 3) y – 2y – 3 = (y + 3)(y – 3)

1 – 2

(y + 3)(y – 3)

(y + 3)(y – 2) = (y + 3)(y – 3) – 2

y2 + y – 6 = y2 – 9 – 2

y = –5

d) 2

x2 – 4 = 1x2 +

1x2 – 2x

2

(x + 2)(x – 2) = 1x2 +

1x(x – 2)

x2(x + 2)(x – 2) 2

(x + 2)(x – 2) = x2(x + 2)(x – 2)

1

x2 + 1

x(x – 2)

2x2 = (x + 2)(x – 2) + x(x + 2)

2x2 = x2 – 4 + x2 + 2x

2x2 = 2x2 + 2x – 4

–x = –2

x = 2

no solution

Solving For a Term in a Formula Solving for any term in a formula is similar Remember: You can always simplify any side of the equation at any time. 1. Simplify the expressions on both sides of the equation. You must

first eliminate all of the Grouping symbols and simplify if necessary.

2. If there are fractions, you may multiply both sides of the equation

by the LCD. Simplify if necessary. 3. You want all of the terms with the desired variable on one side of

the equation and all other term on the other side. If this is not the case, you should Add or subtract to both sides of the equation various variable terms and/or a numerical value in order to get desired variable terns on one side and everything else on the other side. Simplify if necessary.

4. You want one variable term with a coefficient of 1. If this is not

the case, you might have to factor out the desired variable and Divide both sides of the equation by the resulting coefficient. Simplify if necessary.

Note: Always make sure your final answer has the desired variable on the left side of the equation!!!

Examples: Solve the following and use mathematics writing style:

Ex 1: Solve P = 2l + 2w for w

P = 2l + 2w

P – 2l = 2w

P – 2l

2 = w

w = P – 2l

2

Ex 2: Solve A = P(1 + rt) for r

A = P(1 + rt)

A = P + Prt

A – P = Prt

A – P

Pt = r

r = A – P

Pt

Ex 3: Solve A = 2πrh + 2πr2 for h

A = 2πrh + 2πr2

2πrh = A – 2πr2

h = A – 2πr2

2πr

Ex 4: Solve A = h2 (b + B) for B

A = h2 (b + B)

2Ah = b + B

2Ah – b = B

B = 2Ah – b

Ex 5: Solve y = x + 1x – 3 for x

y = x + 1x – 3

y(x – 3) = x + 1

xy – 3y = x + 1

xy – x = 3y + 1

x(y – 1) = 3y + 1

x = 3y + 1 y – 1

Solving Absolute Value Equations: Always put the absolute value on a side by itself if possible!!! 1. If | x | = c (c > 0), then x = c or x = –c 2. If | x | = 0 , then x = 0 3. If | x | = –c (c > 0), then there is no solution If you have an absolute value on both sides: 4. If | x | = | c |, then x = c or x = –c Note: Always make sure your final answer has the variable on the left side of the equation!!!

Solve the following and use mathematics writing style:

Ex 1: | 2x – 5 | = –3

no solution

Ex 2: | x + 4 | = 0

x + 4 = 0

x = –4

Ex 4: | x + 2 | = 3

x + 2 = 3 x + 2 = –3

x = 1 x = –5

x = 1, –5

Ex 6: | 2x + 1 | – 1 = 4

| 2x + 1 | = 5

2x + 1 = 5 2x + 1 = –5

2x = 4 2x = –6

x = 2 x = –3

x = 2, –3

Ex 3: | 12 x – 1 | = 4

12 x – 1 = 4

12 x – 1 = –4

12 x = 5

12 x = –3

x = 10 x = –6

x = 10, –6

Ex 5: | 2 – x | = 1 + x

2 – x = 1 + x 2 – x = –(1 + x)

–2x = –1 2 – x = –1 – x

x = 1/2 0 = –3

x = 1/2

Ex 7: | 3x + 1 | = | x – 1 |

3x+1 = x–1 3x+1 = –(x–1)

2x = –2 3x+1 = –x+1

x = –1 4x = 0

x = 0

x = –1, 0

Solving Word Problems ( Super Solvers Use C.A.P.E.S.)

1. Read the problem carefully. (Reread it several times if necessary)

2. Categorize the problem type if possible. (Is it a problem of numerical expression, distance–rate–time, cost–profit, or simple interest type?)

3. Decide what is asked for, and Assign a variable to the unknown quantity. Label the variable so you know exactly what it represents.

4. Draw a Picture, diagram, or chart whenever possible!!

5. Form an Equation (or inequality) that relates the information provided.

6. Solve the equation (or inequality). 7. Check your solution with the wording of the problem to be sure

it makes sense. distance–rate–time: distance = rate • time (d = r • t) cost–profit: profit = revenue – cost (P = R – C) simple interest: interest = principal * rate (i = p • r)

Example: A grocery store bought ice cream for 59¢ a half gallon and stored it in two freezers. During the night, one freezer "defrosted" and ruined 14 half gallons. If the remaining ice cream is sold for 98¢ a half gallon, how many half gallons did the store buy if it made a profit of $42.44? charge = price • quantity bought .59 x sold .98 x – 14 Example: Last summer, Ernie sold surfboards. One style sold for $70 and the other sold for $50. He sold a total of 48 surfboards. How many of each style did he sell if the receipts from each style were equal? charge = price • quantity one style 70 x other style 50 48 – x

Example: Sellit Realty Company gets a 6% fee for selling improved properties and 10% for selling unimproved land. Last week, the total sales were $220,000 and the total fees were $14,000. What were the sales from each of the two types of properties? fees = rate • sales improved .06 x unimproved .10 220000 – x Example: Maria jogs to the country at a rate of 10 mph. She returns along the same route at 6 mph. If the total trip took 1 hour 36 minutes, how far did she jog? distance = rate • time going x 10 coming x 6

Properties of Exponents Property: Example: b0 = 1 50 = 1

b–n = 1bn 3–2 =

132 =

19

bm • bn = bm + n 23 •25 = 28

bm

bn = bm – n 59

57 = 52 = 25 (bm)n = bm • n (23)4 = 212 = 4096 (a • b)n = an • bn (4x)3 = 43 • x3 = 64x3

a

b n =

an

bn

3

5 2 =

32

52 = 9

25

Some Additional Properties

a

b –n

= bn

an 1

b–n = bn a–n

b–m = bm

an

Examples: Simplify the following using mathematics writing style:

a) x–5 • x9

x3 • x

= x–5+9

x3+1

= x4

x4

= 1

b)

z3z0

zkz –5

=

z3

zk+1 –5

= (z2–k)–5

= z–10+5k

= z5k–10

c)

y–2

y–5 –2

= (y–2+5)–2

= (y3)–2

= y3•(–2)

= y–6

= 1y6

d)

w2w–1

w5w2 –2

=

w1

w7 –2

= (w–6)–2

= w12

e) (x–2y–3)4 (x4y–2)–2

(x3y–1)–3 (x–4y)4

= x–8y–12 x–8y4

x–9y3 x–16y4

= x–16y–8

x–25y7

= x9y–15

= x9

y15

Multiplying Polynomials The product of two polynomials can be found by using some form of the distributive property

a(b + c) = ab + ac. Example: Multiply the given polynomials and simplify. a) (x + 4)(2x – 3) = x(2x – 3) + 4(2x – 3) = 2x2 – 3x + 8x – 12

= 2x2 + 5x – 12 c) (2x – 3y)(x + 2y) = 2x(x + 2y) – 3y(x + 2y) = 2x2 + 4xy – 3xy – 6y2

= 2x2 + xy – 6y2

b) (3x – 5)(x – 2) = 3x(x – 2) – 5(x – 2) = 3x2 – 6x – 5x + 10

= 3x2 – 11x + 10 d) (z – 1)(z – 2) = z(z – 2) – 1(z – 2) = z2 – 2z – z + 2

= z2 – 3z + 2

e) (x + 2y – 3)(2x – y – 2) = x(2x – y – 2) + 2y(2x – y – 2) – 3(2x – y – 2) = 2x2 – xy – 2x + 4xy – 2y2 – 4y – 6x + 3y + 6

= 2x2 – 2y2 + 3xy – 8x – y + 6 f) (x + 2)(x2 – 2x + 4) = x(x2 – 2x + 4) + 2(x2 – 2x + 4) = x3 – 2x2 + 4x + 2x2 – 4x + 8

= x3 + 8

Special Products of Polynomials Take time and learn these formulas. You will need them later and you will be tested on them! (F + S)(F – S) = F2 – S2 Difference of squares (F + S)2 = F2 + 2FS + S2 Perfect square trinomial (F – S)2 = F2 – 2FS + S2 Perfect square trinomial (F + S)(F2 – FS + S2) = F3 + S3 Sum of cubes (F – S)(F2 + FS + S2) = F3 – S3 Difference of cubes

W A R N I N G : Unless x or y is zero, (x + y)2 does not equal x2 + y2!! For example (3 + 4)2 = 72 = 49, but 32 + 42 = 9 + 16 = 25. (Note that 25 is less than 49 by 2(3)(4) = 24, the missing middle term.)

x

x

y

y

x y

x yx2

y2

Table to demonstrate that (x + y)2 = x2 + 2xy + y2 Nonsense like thinking that (x + y)2 equals x2 + y2 will not be tolerated!!

We use the table below for reference to help with multiplication.

(F + S)(F – S) = F2 – S2 Difference of squares (F + S)2 = F2 + 2FS + S2 Perfect square trinomial (F – S)2 = F2 – 2FS + S2 Perfect square trinomial Example: Perform the indicated products and simplify.

F) (x + 3y)(x – 3y) = (F + S)(F – S)

= F2 – S2

= x2 – 9y2

c) (6w2 + 7v5)(6w2 – 7v5) = (F + S)(F – S)

= F2 – S2

= 36w4 – 49v10 e) (4x – 3y)2 = (F – S)2

= F2 – 2FS + S2 = 16x2 – 24xy + 9y2

g) (x3 + 8y)2 = (F + S)2 = F2 + 2FS + S2

= x6 + 16x3y + 64y2

S) (2xy – 5z)(2xy + 5z) = (F – S)(F + S)

= F2 – S2

= 4x2y2 – 25z2 d) (x – 4)2 = (F – S)2 = F2 – 2FS + S2

= x2 – 8x + 16 f) (7w + 5)2 = (F + S)2 = F2 + 2FS + S2

= 49w2 + 70w + 25 h) (2x2z – 3y4)2

= (F – S)2

= F2 – 2FS + S2 = 4x4z2 – 12x2y4z + 9y8

Special Products: Perfect Square Polynomials

Complete the table:

F S F2 2FS S2 (F + S)2 = F2 + 2FS + S2

x 3 (x + 3)2 = _______________

z 4 (z + 4)2 = _______________

w 5 (w + 5)2 = _______________

x –2 (x – 2)2 = _______________

y –1 (y – 1)2 = _______________

z –6 (z – 6)2 = _______________

2x 3y (2x + 3y)2 = _______________

3x –5y (3x – 5y)2 = _______________

4x –y (4x – y)2 = _______________

2x2 5yz3 (2x2 + 5yz3)2 = _______________

Special Products: Difference of Squares

Complete the table:

F S F2 S2 (F + S)(F – S) = F2 – S2

x 3 (x + 3)(x – 3) = _______________

z 4 (z + 4)(z – 4) = _______________

w 5 (w + 5)(w – 5) = _______________

x 2 (x – 2)(x + 2) = _______________

y 1 (y – 1)(y + 1) = _______________

z 6 (z – 6)(z + 6) = _______________

2x 3y (2x + 3y)(2x – 3y) = _______________

3x 5y (3x – 5y)(3x + 5y) = _______________

4x y2 (4x – y2)(4x + y2) = _______________

2x2 5yz3 (2x2+5yz3)(2x2–5yz3) = _______________

FACTORING Factoring is the inverse operation to multiplying polynomials. Remember: In order to be good at factoring polynomials, you must be good at multiplying polynomials. Make sure you are very good at multiplying polynomials and using the special products.

Special Polynomial Factors Common factor ax + ay = a(x + y)

Sum/Diff. of squares x2 + y2

x2 – y2 = does not factor

(x – y)(x + y)

Sum/Diff. of cubes x3 + y3

x3 – y3 = (x + y)(x2 – xy + y2)(x – y)(x2 + xy + y2)

Perfect square trinomial x2 + 2xy + y2

x2 – 2xy + y2 = (x + y)2

(x – y)2

Common Factors & Negative Exponents Anytime you factor, remember to take out the common factor first!! Also, the biggest factor uses the smallest exponent. Examples: Factor the following and use mathematics writing style:

a) x2 + 5x

= x (x + 5)

c) 12x3y2 – 32x2y3 + 20xy4

= 4xy2 (3x2 – 8xy + 5y2)

e) x–5 + 5x–7

= x–7 (x2 + 5)

g) x–3y2 – x–4y6

= x–4y2 (x – y4)

b) 8z7 + 12z4

= 4z4 (2z3 + 3)

d) 9a9b4 – 12a8b6 – 6a7b8

= 3a7b4 (3a2 – 4ab2 + 2b4)

f) 8x–9 – 6x–8

= 2x–9 (4 – 3x)

h) 12a–6b–8 – 16a–8b–6

= 4a–8b–8 (3a2 – 4b2)

Sum / Difference of Squares & Cubes You are required to determine at a quick glance any special product polynomial.

F2 + S2 does not factor, F3 + S3 = (F + S)(F2 – FS + S2) F2 – S2 = (F – S)(F + S) F3 – S3 = (F – S)(F2 + FS + S2)

Examples: Is the polynomial is the sum of squares, difference of squares, difference of cubes, or the sum of cubes? If so, then factor it, if possible.

a) A2 + 64 s.s. does not factor

b) w2 – 25 d.s. (w – 5)(w + 5)

c) z2 – 8 no

d) z3 – 8 d.c (z – 2)(z2 + 2x + 4)

e) 4x2 – 9y4 d.s (2x – 3y2)(2x + 3y2)

f) z3 + 125 s.c. (z + 5)(z2 – 5z + 25)

g) B2 + 9 s.s. does not factor

h) c2 – 16 d.s. (c – 4)(c + 4)

i) c3 – 16 no

j) x2y6 – 81z10 d.s. (xy3 – 9z5)(xy3 + 9z5)

k) 8y3z6 + 27w3 s.c. (2yz2 + 3w)(4y2z4 – 6yz2w + 9w2)

Perfect Square Trinomials You are required to determine at a quick glance any special product polynomial. F2 + 2FS + S2 = (F + S)2 F2 – 2FS + S2 = (F – S)2 Examples: Is the polynomials a perfect square trinomial? If so, then factor it.

x2 – x + 1 no

x2 – 2x + 1 yes (x – 1)2

x2 + 6x + 9 yes (x + 3)2

x2 – 4x – 4 no

x2 – 4x + 4 yes (x – 2)2

a2 – 10a + 25 yes (a – 5)2

9B2 – 12BC + 4C2 yes (3B – 2C)2

z2 + 13z + 36 no

25z2 + 60z + 36 yes (5z + 6)2

Factoring By Grouping Examples: Factor the following and use mathematics writing style:

a) ax + ay + bx + by

= a(x + y) + b(x + y)

= (a + b) (x + y)

c) x2 + 4xy – 2x – 8y

= x(x + 4y) – 2(x + 4y)

= (x – 2)(x + 4y)

e) x3 – 4x2 + 3x – 12

= x2(x – 4) + 3(x – 4)

= (x2 + 3)(x – 4)

b) ax + ay – x – y

= a(x + y) – 1(x + y)

= (a – 1) (x + y)

d) xy – 3x – 4y + 12

= x(y – 3) – 4(y – 3)

= (x – 4)(y – 3)

f) 2x2 – 7xy + 6y2

= 2x2 – 4xy – 3xy + 6y2

= 2x(x – 2y) – 3y(x – 2y)

= (2x – 3y)(x – 2y)

Factoring Polynomials Using The ac–Method

ax2 + bx + c We assume that a, b, and c are integers. Step 1. Multiply a • c Step 2. If you can, find two integers such that: a. their product is ac

b. their sum is b Step 3. If two integers don't exist, STOP because the problem cannot

be factored. Otherwise, move on to Step 4. Step 4. Rewrite the middle term (bx) using the two integers from

Step 2 as coefficients. Step 5. Factor by grouping (like when there are 4 terms). Examples: Factor the following using the ac–method:

a) 4x2 – x – 18 = 4x2 – 9x + 8x– 18 = x(4x – 9) + 2(4x – 9)

= (x + 2)(4x – 9)

b) 12x2 – 23x – 24 = 12x2 – 32x + 9x– 24 = 4x(3x – 8) + 3(3x – 8)

= (4x + 3)(3x – 8)

Factoring With Substitution Examples: Factor the following and use mathematics writing style:

a) x2y6 – 81z10

= F2 – S2 = (F – S)(F + S)

= (xy3 – 9z5)(xy3 + 9z5)

c) x2 – (y + 2)2

= F2 – S2 = (F – S)(F + S) = [x – (y + 2)][x + (y + 2)]

= (x – y – 2)(x + y + 2)

e) (2x – 5)2 – 2(2x – 5) – 8

= F2 – 2F – 8 = (F – 4)(F + 2) = [(2x – 5) – 4][(2x – 5) + 2] = [2x – 5 – 4][2x – 5 + 2]

= (2x – 9)(2x – 3)

b) x3 + 8

= F3 + S3

= (F + S)(F2 – FS + S2)

= (x + 2)(x2 – 2x + 4)

d) (x + 2y)2 – 25

= F2 – 25 = (F – 5)(F + 5) = [(x + 2y) – 5][(x + 2y) + 5]

= (x + 2y – 5)(x + 2y + 5)

f) (x – 3)2 – 9(y + 2)2

= F2 – 9S2 = (F – 3S)(F + 3S) = [(x–3) –3(y+2)][(x–3) +3(y+2)] = [x –3 – 3y –6)][x –3 + 3y + 6]

= (x – 3y – 9)(x + 3y + 3)

Factoring Completely You are always required to continue to factor until every expression can be factored no further.

Steps For Factoring Polynomials 1. Always take out the greatest common factor first!!! (This includes the case with negative exponents.) 2. See if you can use one of the special factors. Check how many

terms are present. A. If there are 2 terms: a. See if it is a difference of squares b. See if it is a sum or difference of cubes B. If there are 3 terms: See if it is a perfect square trinomial C. If there are 4 terms: See if you can factor by grouping. 3. Try factoring by any other method you can. (Trial and error, ac method, etc.) Hint: Whenever you have to factor, DO NOT MULTIPLY out the terms unless you absolutely have to !! Consider using a substitution.

Examples: Factor completely the following and use mathematics writing style:

a) w–7 – 4w–9

= w–9(w2 – 4)

= w–9(w – 2)(w + 2)

c) (x2 – x)2 – 18(x2 – x) + 72

= F2 – 18F + 72

= (F – 6)(F – 12)

= (x2 – x – 6)(x2 – x – 12)

= (x – 3)(x + 2)(x – 4)(x + 3)

b) 4 + 4z–1 + z–2

= z–2(4z2 + 4z + 1)

= z–2(2z + 1)2

d) x8 – y8

= (x4 – y4)(x4 + y4)

= (x2 – y2)(x2 + y2)(x4 + y4)

= (x–y)(x+y)(x2+y2)(x4+y4)

e) (x2 + 3x – 10)2 – (x – 2)2

= F2 – S2

= (F + S)(F – S)

= [(x2 + 3x – 10) + (x – 2)][(x2 + 3x – 10) – (x – 2)]

= [x2 + 3x – 10 + x – 2][x2 + 3x – 10 – x + 2]

= (x2 + 4x – 12)(x2 + 2x – 8)

= (x + 6)(x – 2)(x + 4)(x – 2)

= (x + 6)(x + 4)(x – 2)2

Basic Properties of Rational Expressions

A rational expression is any expression of the form

PQ

where P and Q are polynomials and Q ≠ 0. In the following properties, no denominator is allowed to be zero. Basic Property: Example:

PQ =

RS if and only if P • S = Q • R

23 =

69

PQ =

P • RQ • R

35 =

3 • 45 • 4 =

1220

– PQ = –PQ =

P–Q = –

–P–Q – 74 =

–74 =

7–4 = –

–7–4

PQ = –

–PQ = –

P–Q =

–P–Q

58 = –

–58 = –

5–8 =

–5–8

Notice that

–PP = –1.

Since it is true that x – a = – (–x + a) = – (a – x), It must also be true that

x – aa – x = –1.

W A R N I N G : You must reduce only factors!! If the terms are not factors, they cannot be factored out. Nonsense like "canceling out" nonfactors will not be tolerated!!

2x + 8

2 = x + 4

x2 – 9x – 3 = x + 3

Multiplication and Division of Rational Expressions

To multiply two rational expression, multiply the numerators together and multiply the denominators together.

PQ •

RS =

P • RQ • S .

To divide two rational expressions, invert the one immediately after the division sign, and do a rational expression multiplication.

PQ ÷

RS =

PQ •

SR =

P • SQ • R

Example: Perform the indicated operation and simplify:

x2 + 2x – 3x2 – 3x – 10 ÷

4x + 2x2 – x •

2x2 – 9x – 5x2 – 2x + 1

= x2 + 2x – 3x2 – 3x – 10 •

x2 – x4x + 2 •

2x2 – 9x – 5x2 – 2x + 1

= (x + 3)(x – 1)(x – 5)(x + 2) •

x(x – 1)2(2x + 1) •

(2x + 1)(x – 5)(x – 1)2

= x(x + 3)2(x + 2)

Rule for Adding or Subtracting Fractions with Equal Denominators

(RASFED) To add or subtract two rational expressions whose denominators are equal, simply add or subtract the numerators. Make sure you use parentheses when appropriate!

PQ +

RQ =

P + RQ

PQ –

RQ =

P – RQ

Examples: Perform the indicated operation and simplify:

a) x

x2 – 4 + 2

x2 – 4

= x + 2x2 – 4

= x + 2

(x + 2)(x – 2)

= 1

x – 2

b) x2

x2 – 4 – 4x – x2

x2 – 4

= x2 – (4x – x2)

x2 – 4

= x2 – 4x + x2

x2 – 4

= 2x2 – 4x

x2 – 4

= 2x(x – 2)

(x + 2)(x – 2)

= 2x

(x + 2)

Finding the LCD Step 1. Factor each denominator completely, including the prime

factors of any constant factor. Step 2. Form the product of all the factors that appears in the

complete factorizations. Step 3. The number of times any factors appears in the LCD is the

most number of times it appears in any one factorization. Examples: Find the LCD for the given denominators: a) Denominators are 24, 30, and 36 24 = 8 • 3 = 23 • 3 30 = 6 • 5 = 2 • 3 • 5 36 = 4 • 9 = 22 • 32 LCD = 23 • 32 • 5 = 360 b) Denominators are x3 – x2 and x3 – x x3 – x2 = x2(x – 1) x3 – x = x(x2 – 1) = x(x + 1)(x – 1) LCD = x2(x + 1)(x – 1)

Rule for Adding or Subtracting Fractions with Unequal Denominators

(FLEAS)

1. Factor the rational expression.

2. Find the Least Common Denominator (LCD).

3. Equalize each denominators by replacing each fraction with an equivalent one whose denominator is the LCD.

4. Add or Subtract using RASFED. Example:

12 –

23 +

34

= 6

12 – 8

12 + 9

12

= 6 – 8 + 9

12

= 7

12

Examples: Perform the indicated operation and simplify.

a) 2

x2 – 1 – 1

x2 + 2x + 1

= 2

(x + 1)(x – 1) – 1

(x + 1)2

= 2(x + 1)

(x + 1)2(x – 1) – x – 1

(x + 1)2(x – 1)

= 2(x + 1) – (x – 1)

(x + 1)2(x – 1)

= 2x + 2 – x + 1(x + 1)2(x – 1)

= x + 3

(x + 1)2(x – 1)

b) x + 2x – 2 –

x2 + 2xx2 – 4

= x + 2x – 2 –

x2 + 2x(x + 2)(x – 2)

= (x + 2)2

(x + 2)(x – 2) – x2 + 2x

(x + 2)(x – 2)

= (x2 + 4x + 4) – (x2 + 2x)

(x + 2)(x – 2)

= x2 + 4x + 4 – x2 – 2x

(x + 2)(x – 2)

= 2x + 4

(x + 2)(x – 2)

= 2(x + 2)

(x + 2)(x – 2)

= 2

(x – 2)

Complex Fractions A simple fraction is any rational expression whose numerator and denominator contain no rational expression. A complex fraction is any rational expression whose numerator or denominator contains a rational expression.

To simplify complex fractions: Step 1: Identify all fractions in the numerator and denominator and

find the LCD. Step 2: Multiply the numerator and denominator by the LCD. Examples:

a)

12 –

13

34 –

16

= 12

1

2 – 13

12

3

4 – 16

= 6 – 49 – 2

= 27

b) x + y

x–1 + y–1

= x + y1x +

1y

= xy (x + y)

xy

1

x + 1y

= xy (x + y)

y + x

= xy

Example: Simplify the following:

a) 9 –

1y2

3 – 1y

= y2

9 – 1y2

y2

3 – 1y

= 9y2 – 13y2 – y

= (3y + 1)(3y – 1)

y(3y – 1)

= 3y + 1

y

b)

1x + h –

1x

h

= x(x + h)

1

x + h – 1x

x(x + h) h

= x – (x + h)x(x + h) h

= x – x – hx(x + h) h

= –h

x(x + h) h

= –1

x(x + h)

Rational Exponents If a is a real number and n is a positive integer, we sometimes speak of the nth root of a. If it exists, it should be some number b whose nth power is equal to a. That is, we should have bn = a, and then it seems reasonable to write b = a1/n. If a is a positive real number and n is a positive integer, the principal nth root of a is the positive real number whose nth power is equal to a. We denote the principal nth root of a by using

1n as an exponent and we write

a1/n .

To avoid any ambiguity, use the principal nth root of a to calculate a1/n.

If a is any real number and n is a positive integer:

a > 0 a < 0 a = 0

a1/n is a positivereal number

a1/n = 0n odd n even

a1/n is a negativereal number

a1/n is not areal number

More Rational Exponents If a is any real number, m any nonzero integer, and n a positive integer such that a1/n is defined. Then we have am/n = (a1/n)m = (am)1/n To calculate am/n, you should take the root first!! am/n = (a1/n)m .

Radicals If a is a real number, and n is a positive integer such that a1/n is defined, then we have

n a = a1/n. Thus we also have

am/n =

n a

m = n am

Again, to calculate am/n, you should take the root first!!

am/n =

n a

m .

Examples: Simplify the following expressions.

a) 161/2

= 4

b) 1251/3

= 5

c) 811/4

= 3

d)

16

25 1/2

= 45

e)

27

8 1/3

= 32

f)

–1

32 1/5

= –12

g) 84/3

= (81/3)4

= 24

= 16

h) 8–2/3

= (81/3)–2

= 2–2

= 14

i) (–4)–1/2

not real

j) 25–3/2

= (251/2)–3

= 5–3

= 1

125

k) –25–3/2

= –(251/2)–3

= –5–3

= –1

125

l) 3 • 16–5/4

= 3 • (161/4)–5

= 3 • 2–5

= 3 • 1

32

= 3

32

Examples: Simplify the following expressions. Assume that all variables are positive.

a) (2x1/3)3

= 23 (x1/3)3

= 8x

b) y2/7 • y3/7

= y(2/7 + 3/7)

= y5/7

c) 4z1/2 • 3z1/3

= 12z(1/2 + 1/3)

= 12z5/6

d) x2/3

x1/4

= x(2/3 – 1/4)

= x5/12

e) (x1/4y1/2)3

x1/2y–1/4

= x3/4y3/2

x1/2y–1/4

= x(3/4 – 1/2)y(3/2 + 1/4)

= x1/4y7/4

Properties of Radicals Remember that if no index is shown, then it is by default 2. Thus 2 x means x . The properties of radicals come directly from the definition and the properties of exponents. In particular, we have Property Example

n a • b = n a • n b 3 8x12 = 3 8 • 3 x12 = 2x4

n a

b = n an b

4 16

81 = 4 164 81

= 23

m

n a = mn a 5

x = 10 x

Absolute Value Again The absolute value can be defined in terms of the square root:

| x | = x2

Simplifying Radicals A radical expression is considered to be simplified when:

1. The radicand has no factors with an exponent greater than or equal to the index of the radical, 3 x6y11 = 3 x6y9y2 = 3 x6y93 y2 = x2y3 3 y2 2. No radical contains a fraction,

3 32x4 =

3 12x2

8x6 = 3 12x2

2x2 3. No denominator contains a radical, and

21 – 3 x =

2(1 + 3 x)(1 – 3 x)(1 + 3 x) =

2(1 + 3 x)1 – 9x

4. The index and the power of the radicand have no common factor except 1. 9 x6 = 3 x2

Examples: Simplify the following expressions. Assume that all variables are positive.

a) 98

= 49 • 2

= 49 2

= 7 2

b) 3 56

= 3 8 • 7

= 3 8 3 7

= 23 7

c) 4 243

= 4 81 • 3

= 4 81 4 3

= 34 3

d) 5

4x

= 5x4x2

= 5x

2x

e) 4 16x

9y

= 4 16x•9y3

81y4

= 24 9xy3

3y

g) 3

4 x

= 12 x

f) 3

3 32x

= 3

3 32x •

3 2x2

3 2x2

= 33 2x2

3 64x3

= 33 2x2

4x

h) 4 3x5

8y7

= 4 3x5 • 2y

8y7 • 2y

= 4 6x5y

16y8

= x4 6xy

2y2

Examples: Simplify the following expressions. Assume that all variables are positive.

a) 2 10 – 6 10

= –4 10

b) 32x + 18x

= 16•2x + 9•2x

= 4 2x + 3 2x

= 7 2x

c) 12 + 18 + 27

= 4•3 + 9•2 + 9•3

= 2 3 + 3 2 + 3 3

= 5 3 + 3 2

d) 5 + 45 – 15

= 5 + 9•5 – 15

= 5 + 3 5 – 15

= 4 5 – 15

e) 8x – 3 2x + 18x

= 4•2x – 3 2x + 9•2x

= 2 2x – 3 2x + 3 2x

= 2 2x

f) 3 5x – 3 40x

= 3 5x – 3 8•5x

= 3 5x – 23 5x

= –3 5x

g) 4 x7 + x4 81x3

= 4 x4•x3 +x4 81•x3

= x4 x3 + 3x4 x3

= 4x4 x3

h) 5 x5y10z + 4xy25 z

= xy25 z + 4xy25 z

= 5xy25 z

i) ( 7 – 2)( 7 + 3)

= ( 7 )2 + 3 7 – 2 7 – 6

= 7 + 7 – 6

= 1 + 7

j) ( 6 – 2 5 )( 6 + 5 )

= 6 + 30 – 2 30 – 2•5

= 6 – 30 – 10

= –4 – 30

k) ( 2 + 5)( 2 – 5)

= ( 2 )2 – (5)2

= 2 – 25

= –23

l) ( 3 + 2 )2

= 3 + 2 3 2 + 2

= 3 + 2 6 + 2

= 5 + 2 6

m) (3 + 2 )2

= 32 + 2•3 2 + 2

= 9 + 6 2 + 2

= 11 + 6 2

n) ( 6 + 2 )2

= ( 6 )2 + 2 6 2 + ( 2 )2

= 6 + 2 12 + 2

= 6 + 2•2 3 + 2

= 8 + 4 3

o) 2

4 – 3 2

= 2(4 + 3 2)

(4 – 3 2)(4 + 3 2)

= 2(4 + 3 2)16 – 9•2

= 2(4 + 3 2)

16 – 18

= 2(4 + 3 2)

–2

= –(4 + 3 2 )

p) 31

6 + 5

= 31(6 – 5)

(6 + 5)(6 – 5)

= 31(6 – 5)

36 – 5

= 31(6 – 5)

31

= 6 – 5

q) 1

2 7 – 3 2

= 1(2 7 + 3 2)

(2 7 – 3 2)(2 7 + 3 2)

= 2 7 + 3 24•7 – 9•2

= 2 7 + 3 2

28 – 18

= 2 7 + 3 2

10

r) 2 + 53 – 5

= (2 + 5)(3 + 5)(3 – 5)(3 + 5)

= 6 + 5 5 + 5

9 – 5

= 11 + 5 5

4

s) 4 2

2 2 + 3

= 4 2(2 2 – 3)

(2 2 + 3)(2 2 – 3)

= 8•2 – 12 2

4•2 – 9

= 16 – 12 2

8 – 9

= 16 – 12 2

–1

= –16 + 12 2

t) x – 5x – 5

= (x – 5)( x + 5)

( x – 5)( x + 5)

= (x – 5)( x + 5)

x – 5

= x + 5

Graphing Equations Caution: We use parentheses to represent both an ordered pair and an open interval. The context of the discussion should clarify whether points in the plane or intervals is needed. You must know the difference between the two concepts!

!x= 1.00!y= 1.00

5.00

-5.00

-5.00 5.00 x

y

Notice that: 1. Every pair of real number corresponds to a unique point in the

plane and every point in the plane corresponds to a unique pair of real numbers.

2. The plane is divided into four regions called quadrants.

!x= 1.00!y= 1.00

5.00

-5.00

-5.00 5.00 x

y

Graphing Linear Equations In Two Variables

1. Calculate any three points that satisfy the equation. 2. Plot these points in the graph. 3. If a straight line passes through these points, draw the line. If not,

go back to step 1. Example: Graph the lines a) x + y = 2 b) 2x – 3y = 6 c) y = 4 d) x = –3

!x= 1.00!y= 1.00

5.00

-5.00

-5.00 5.00 x

y

Example: Graph the following (nonlinear) equations: a) y = x2 b) y = x2 – 4

!x= 1.00!y= 1.00

5.00

-5.00

-5.00 5.00 x

y

!x= 1.00!y= 1.00

5.00

-5.00

-5.00 5.00 x

y

c) y = | x | d) y = | x | + 1

!x= 1.00!y= 1.00

5.00

-5.00

-5.00 5.00 x

y

!x= 1.00!y= 1.00

5.00

-5.00

-5.00 5.00 x

y

Introduction to Functions

A relation is a correspondence between two sets that assigns to each element of the first set an element of the second set. For us, we will consider a relation to be a set of ordered pairs.

The domain, D, of a relation is the set of all first components in

the relation. The range, R, of a relation is the set of all second components in

the relation.

Example: Find the domain and range for each of the following relations. 1. r = {(2, 3), (4, –3), (–1, 1), (4, 0), (–3, 9)} Solution: D = {–3, –1, 2, 4} (all of the x–coordinates) R = {–3, 0, 1, 3, 9} (all of the y–coordinates) 2. r = {(1, 2), (3, 1), (2, –2), (–5, 1)} Solution: D = {–5, 1, 2, 3} (all of the x–coordinates) R = {–2, 1, 2} (all of the y–coordinates)

A function is a relation in which each domain element has only

one corresponding range element. Example: Determine whether or not each of the following relations is a function. 1. r = {(2, 3), (4, –3), (–1, 1), (4, 0), (–3, 9)} 2. r = {(1, 2), (3, 1), (2, –2), (–5, 1)} 3. r = {(2, 3), (1, 6), (2, 5), (0, –1)} 4. r = {(–1, 1), (1, 5), (0, 3)} Example: Determine whether or not each of the following relations is a function. 1. y = 2x + 1 2. x = y2

3. y = x

x + 3

A function is a relation in which each domain element has only one corresponding range element.

It then follows that a function is a relation in which no two ordered pairs have the same first element. Graphically, this means that if two or more points in a relation have a common vertical line passing through them, then the relation is not a function. Vertical Line Test: If any vertical line intersects a graph of a relation in more than one point, then the relation is not a function. Example: Which of the following graphs is the graph of a function?

a)

!x= 1.00!y= 1.00

5.00

-5.00

-5.00 5.00 x

y

b)

!x= 1.00!y= 1.00

5.00

-5.00

-5.00 5.00 x

y

c)

!x= 1.00!y= 1.00

5.00

-5.00

-5.00 5.00 x

y

d)

!x= 1.00!y= 1.00

5.00

-5.00

-5.00 5.00 x

y

Function Notation Sometimes we describe functions in terms of a formula. To do this, we need three things: 1) The name of the function f 2) The “dummy” or “place holder” variable x 3) The formula that describes the function 2x + 1 For example, we could have the function defined by f(x) = 2x + 1. Then we can calculate other values as follows. f(x) = 2x + 1 f(3) = 2(3) + 1 = 7 f(–7) = 2(–7) + 1 = –13 Please don’t confuse the notation f(x) with multiplication!

Example: Let f(x) = x2 – 2x and g(x) = 1x . Calculate the following:

a) f(4) b) g(6) c) f(–3) d) g(0.25) e) f(2) f) g(0)

Slopes The slope m of the line passes through the points (x1,y1) and (x2,y2) is given

by m = y2 – y1x2 – x1 .

Example: Find the slope of the line passing though the given points: a) (–1, 2) and (3, 5)

m = 5 – 2

3 – (–1)

= 34

b) (0, 1) and (2, 6)

m = 6 – 12 – 0

= 52

c) (5, 1) and (1, 3)

m = 3 – 11 – 5

= 2–4

= – 12

d) (2, –3) and (–1, –9)

m = –9 – (–3)

–1 – 2

= –6–3

= 2

e) (1, 3) and (4, 6)

m = 6 – 34 – 1

= 33

= 1

f) (3, 6) and (1, 6)

m = 6 – 61 – 3

= 0–2

= 0

g) (–3, 2) and (–3, 5)

m = 5 – 2

–3 – (–3)

= 30

slope undefined

Various Linear Equations

Standard Form An equation that can be written in the form

Ax + By = C, (A, B not both zero) is a linear equation in the variables x and y. This form is called the standard form. Intercepts: The x–intercept is the point (if any) where the line crosses the x–axis; and the y–intercept is the point (if any) where the line crosses the y–axis. To find the x–intercept, let y = 0 and solve the equation for x. To find the y–intercept, let x = 0 and solve the equation for y.

Point–Slope Form The equation of the line passing through the point (x1, y1) with slope m is given by

y – y1 = m(x – x1),

and we call this the point–slope form of the equation of the line. Since we need point information and slope information, this is the form used most often to find the equation.

Slope–Intercept Form The equation of the line with slope m and y–intercept b is given by

y = mx + b, and we call this the slope–intercept form of the equation of the line. Given the equation of a line, if we can write the equation of the line into this form (by solving for y), we can determine the slope and the y–intercept of the line.

Horizontal Line Form The equation of the horizontal line that passes through the point (a, b) is given by

y = b.

Vertical Line Form The equation of the vertical line that passes through the point (a, b) is given by

x = a.

Examples: Find the slope and the y–intercept of each of the following lines: a) –2x + 3y = 6 3y = 2x + 6

y = 23 x + 2

slope 2/3 y–intercept 2

b) y = –3x + 2 slope –3 y–intercept 2

c) 5x – 4y = 20 –4y = –5x + 20 y = 54 x – 5 slope 5/4 y–intercept –5

d) 3y + 6 = 0 3y = –6 y = –2 slope 0 y–intercept –2

e) y = 6 slope 0 y–intercept 6

f) x = 3 slope undefined no y–intercept

Finding Equations Of Lines To find the equation of any line, you always need two types of information: slope information and point information.

Point information includes: any point the line passes through y–intercept of the line x–intercept of the line Slope information includes: slope of line two points the line passes through knowing line is horizontal or vertical any parallel or perpendicular line

The basic equations of the line we use are listed:

1. Point–Slope y – y1 = m(x – x1) 2. Slope–Intercept y = mx + b 3. Horizontal line y = b 4. Vertical line x = a

Helpful information to remember:

a. The slope of the line passing through two points is m = y2 – y1x2 – x1

.

b. Slopes of parallel lines are equal. c. Slopes of perpendicular lines are negative reciprocals.

Examples: Find the equation of the line: (Put answers in slope–intercept form if possible) a) through (2, 3) with slope –1/2

y – (3) = – 12 (x – 2)

y – 3 = – 12 x + 1

y = – 12 x + 4

b) through (–4, –1) with slope 2 y – (–1) = 2(x – (–4)) y + 1 = 2(x + 4) y + 1 = 2x + 8

y = 2x + 7

c) through (–1, 2) and (4, –2)

m = –2 – 2

4 – (–1)

= – 45

y – 2 = – 45 (x – (–1))

y – 2 = – 45 (x + 1)

y – 2 = – 45 x –

45

y = – 45 x +

65

d) through (6, –2) and (2, 0)

m = 0 – (–2)

2 – 6

= – 12

y – 0 = – 12 (x – 2)

y = – 12 x + 1

e) through (2, 3) and horizontal

y = 3

f) through (–2, 6) and vertical

x = –2

g) with slope 3 and y–intercept 5

y = 3x + 5

h) with slope 2 and x–intercept 7 y – 0 = 2(x – 7)

y = 2x – 14

i) through (2, 3) parallel to 5x + 3y = 1 find slope: 5x + 3y = 1 y = –5/3 x + 1/3 m = –5/3

y – 3 = – 53 (x – 2)

y – 3 = – 53 x +

103

y = – 53 x +

193

j) through (0, 0) perpendicular to 2x + y = 1 find slope: 2x + y = 1 y = –2x + 1

m = 12

y – 0 = 12 (x – 0)

y = 12 x

k) through (–1, –2) parallel to x + y = 6 find slope: x + y = 6 y = –x + 6 m = –1 y – (–2) = –1(x – (–1)) y + 2 = –1(x + 1) y + 2 = –x – 1

y = –x – 3

l) through (3, 0) perpendicular to x + 2y = 3 find slope: x + 2y = 3 y = –1/2 x + 3/2 m = 2 y – 0 = 2(x – 3)

y = 2x – 6