learning development — birkbeck, university of london

www.mathsupport.wordpress.com Jackie Grant, Birkbeck College, 2012 Creative Commons Copyright : Attribution, Noncommercial, Share Alike

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Algebra 1

In this section we will look at

- the meaning of symbols - the language of algebra - substituting numbers into algebraic expressions - simplifying algebraic expressions - the meaning of equations - rearranging equations - solving equations

Helping you practice

At the end of the sheet there are some questions for you to practice. Don’t worry if you can’t do these but do try to think about them. This practice should help you improve. I find I often make mistakes the first few times I practice, but after a while I understand better.

Videos

All the examples in this worksheet and all the answers to questions are available as answer sheets or videos.

Good luck and enjoy!

Videos and more worksheets are available in other formats from www.mathsupport.wordpress.com


1. The meaning of symbols

Symbols are everywhere. Most people recognise symbols such as

etc. But other symbols are not so familiar

Δ x etc. The important thing is not to panic if you see a new symbol.

When you do see a new symbol, you first need to find out what it means. Often the question or the context will tell you what a symbol represents.

Here is an example. If the symbol means one biscuit, then we can ask how many biscuits Joshua eats in a day. Suppose he has two with his morning tea and 3 during a meeting and 1 at home in the evening. We can write the number of biscuits in this day as

2 + 3 +1

where represents 1 biscuit. This is algebra!

We can then ‘simplify’ this, since 2 biscuits and 3 biscuits and another biscuit is 6 biscuits, so we can write

2 + 3 + 1 = 6

Notice though that it doesn’t matter what symbol we use. As long as we state what the symbol means. So now let’s use the letter ‘a’ to represent one biscuit. Now we write 2 biscuits plus 3 biscuits plus 1 biscuit equals 6 biscuits as

2a + 3a +a = 6a

Notice that ‘a’ on its own is the same as ‘1a’. We could also have used z

2z + 3z + z = 6z

Or any other letter, let’s use the Greek letter theta (pronounced ‘th’ ‘eater’ )

2 + 3 + = 6


Practice and the negative sign

Now we look at examples that include two or more different symbols and we will use the negative sign.

Example 1. 3 friends order breakfast in a cafe. The first asks for 2 sausages, one egg and one piece of toast. The second person asks for 1 sausage, 2 eggs and three pieces of toast and the third person asks for 1 sausage, 1 egg and 3 pieces of toast. We want to calculate the total amount of food

Let’s use S for a sausage, E to represent an egg, and T for one piece of toast

Person 1 orders 2S + 1 E + 1T

Person 2 orders 1S + 2E + 3T

Person 3 orders 1S + 1E + 3T

If we add these all together we get

2S + 1E + 1T + 1S + 2E + 3T + 1S + 1E + 3T

Notice we can simplify this and write the whole breakfast order as

4S + 4E + 7T

Which means the total amount is 4 sausages, 4 eggs, and 7 pieces of toast.

Example 2. Suppose you work for a large company and need to know how many boxes of paper and envelopes are in stock. You count 50 boxes of paper in one office, 20 boxes of paper in the next and 30 boxes in a third office. You also count 7 boxes of envelopes, but someone then takes 4 of the boxes of envelopes and 10 boxes of paper to use.

Using p for each box of paper and e for each box of envelopes we can write

50p + 20p + 30p + 7e – 10p – 4e

Where the – 10p and the – 4e represents the fact that someone is going to use up 10 boxes of paper and 4 boxes of envelopes. Simplifying this gives

90p + 3e

So the total amount is 90 boxes of paper and 3 boxes of envelopes.


Language

Here we will look at the meanings of five key words used in algebra. These are: expression, constant, variable, operation and simplify.

Expression: In algebra we often put different things together in a ‘list’.

50p + 20p +30p + 7e – 10p – 4e

This ‘list’ is called an expression or an algebraic expression. We officially define an expression as a combination of constants, variables and operations.

Constants: are things that do not change (for example the number 30)

Variables: are the symbols that can vary (which means change)or can represent a variety of things (for example E, T, y, are variables and could mean eggs or elephants, or the price of cheese)

Operations: are things you do such as adding or subtracting (+, –, x and )

Simplify: A question might ask you to ‘simplify the algebraic expression’

50p + 20p +30p + 7e – 10p – 4e

in which case we would write

100p + 7e – 10p – 4e = 90p – 3e

Here we have shown each step of our working, and the final line 90p – 3e is the algebraic expression in its simplest possible form.

2. Substitution

Here we introduce anther common word substitution. Substitution is used when we know the value of our variables (or symbols).

Example: if we order four cups of tea and two cakes we can write the cost as

4T + 2C

Where T means the cost of a cup of tea, and C means the cost of a cake. If someone then tells us that a cup of tea is £1 and each cake is £1.50 then we can ‘substitute’ these values into our ‘algebraic expression’.


Remembering that 4T means four lots of tea, or 4 times T or 4 x T, and 2C similarly means 2 cakes or 2 x C. So this substitution can be written as

4T + 2C = 4 x £1 + 2 x £1.50

We can then calculate the total using BIDMAS rules. (If you want to review the BIDMAS rules visit the Numbers section at www.mathsupport.wordpress.com)

4T + 2C = 4 x £1 + 2 x £1.50 = £4 + £3 = £7

We have also shown each step of our working here.

3. More on simplifying

We will now start to use all the operations of BIDMAS with algebra. Remember that BIDMAS tells you the order of each operation. It is short for Brackets, Indices, Division and Multiplication, Addition and Subtraction.

Example 1: There are 3 tables of customers in a cafe. The people at the first table order 2 lasagnes and one salad. The second table orders 3 lasagnes, each with chips and the third table orders 2 salads, each with chips. Let’s use L for lasagne, S for salad and C for chips. We can then write

Table 1 orders 2L + S and table 2 orders 3(L + C) and table 3 orders 2(S + C)

Remember the brackets in 2(S + C) mean that there are two lots of salad and two lots of chips. We can add these together to get an expression for the total

2L + S + 3 (L+C) + 2 (S+C)

Since whatever is outside the bracket multiplies everything inside the brackets, 2(S + C) means 2x(S + C), which is 2S + 2C. So multiplying out brackets gives

2L + S + 3L + 3C + 2S + 2C

This can then be simplified further by combining all the Ls, Ss and Cs to get

5L + 3S + 5C

This is the same as the original expression, but now in its simplest form.

Here are some more worked examples. Since we are practicing simplifying now, so we will just consider the expressions without explaining the symbols.


Example 2: Write the following expression in its simplest form

4z – 8y – 8(z – 2 y)

We start with the brackets (remember the order is BIDMAS)

4z – 8y – 8z + 16y

Notice that the – 8 outside the bracket multiplies both the terms in the brackets, and since negative multiplied by negative gives positive, we have + 16y. There are no indices, multiplication or division that can be simplified, so we add and subtract to get

– 4z + 8y

Example 3: Here we include indices (remember 52 = 5 x 5, so z2 = z x z )

4z2 + 8y – 8(z + 2y)

Again we want to simplify and starting with multiplying out the brackets we get

4z2 + 8y – 8z – 16y

We have to be careful because z2 is different to z. So 4z2 and – 8z cannot be added. But we can add the 8y and the – 16y. So the simplified expression is

4z2 – 8z – 8y


4y2 + 2y – 2(y – 4) – 8 + (4y)2

There is a lot in this expression so we need to be careful. Start by asking if we can simplify the brackets. The first brackets – 2(y – 4) simplifies to – 2y + 8 and the second (4y)2 is the same as 42y2 which is 16y2. So we can write

4y2 + 2y – 2y + 8 – 8 + 16y2

We can only add or subtract y with other y (we can’t add or subtract y with an y2). So the 2y – 2y gives zero and 8 – 8 gives zero and we get the simplest version of the original expression as

20y2


Example 5: Write the following expression in its simplest for

This expression has a lot in it. But don’t panic.

We can consider this expression to be in three sections (these sections are called ‘terms’). The first term is a fraction and we look to see if anything on the top of the fraction cancels with anything on the bottom of the fraction.

To start we write out the first term in full to help us simplify:

were we have cancelled the common terms (in red). To revise cancelling fractions visit the Numbers section at www.mathsupport.wordpress.com

Simplifying the second term gives:

where we have just multiplied out brackets. The third term cannot be simplified, so we leave it as it as and now combine all three terms to get

This can be simplified further because we have four lots of plus another one, and three minus another one. So the simplest form is


Here we multiply out the brackets and simplify, showing our workings as

Practice is the key: Like all things, if you practice it will help your understanding. There are exercises below to help you (answers will full workings are available) so don’t worry if you don’t get the right answers first time.


4. Equations

Here we start dealing with equations. Equations are two algebraic expressions linked by an equals sign. The equal sign means that the two expressions have the same value, in other words they balance. Here are some examples

Equation example 1:

T = 6.00

This could mean that the price of a t-shirt (T) is £6.00, or the time (T) is 6.00am. It is the simplest type of equation as it has one variable T on the left-hand side and the value for it on the right-hand side.

Equation example 2:

2T – y = 10

Here we have two variables T and y. This could mean that the price of 2 t-shirts minus a discount y is £10. We can’t tell the value of T or y from this equation on its own, but if we know either T or y we can calculate the other value

Equation example 3:

E = mc2

This is a famous equation first written by Einstein who used the symbols E for energy, m for mass and c for the speed of light.

Equation example 4:

This is the equation used for calculating the average ( ) of 4 values of . Where the first value is and the second valued and so on. This equation and others like it are used a lot in statistics, and will be covered in the Statistics section at www.mathsupport.wordpress.com

Key point: The important thing about an equation is that there are two expressions (one on the right-hand side and one on the left-hand side). These expressions are linked by the equals sign and so both expressions have the same value.


2T – y 10 equals

5. Re-arranging and solving equations

Because the expressions on each side of an equation are equal, we can say they are balanced. As long as we keep the equation balanced we can move the variables, constants and operations around to find different things.

It is very useful to re-arrange equations and we will see why in a minute.

One way to picture this re-arranging and balancing is to imagine a see-saw. If the see-saw is level then left-hand side balances with the right-hand side. Using the equation 2T – y = 10 we can draw this balancing as

We can now follow the BIDMAS rules and do anything to the see-saw as long as we do the same to both sides as this will keep the see-saw balanced. Keeping the see-saw balanced means that we keep everything on the right-hand side equal to everything on the left-hand side: so the equation is still valid.

In the following steps we are going to add and subtract using the BIDMAS rules to re-arrange the equation to find y. We need to add and subtract things to get the y on its own and positive, there are many ways to do this. Here is one way

Step 1: Since y is negative and on the left-hand side, we can remove it from the left-hand side by adding y to both sides, this would give us

2T – y + y = 10 + y

Step 2: Since – y + y is zero, simplifying the expression gives

2T = 10 + y

Step 3: We now have y + 10 on the right-hand side and if we remove the 10 we will have y on its own, which is what we want. To remove the 10 which is adding, we need to do the opposite, which is subtracting. So we subtract 10 from the right-hand side, but to keep the see-saw balanced we have to do this to both sides

2T – 10 = 10 + y – 10


Step 4: Since 10 + y – 10 is just y , simplifying gives

2T – 10 = y

Step 5: Since both sides are still equal we can swap them over, giving

y = 2T – 10

So we started out with 2T – y = 10 and re-arranged to get y = 2T – 10. Both equations represent the same thing, but in the final version ‘y is the subject’.

Example: Re-arrange the following to make z the subject

y = 3z +7

We want to remove everything from around the z on the right-hand side, so that just the z remains on its own.

Step 1: Remove the +7. To do this we subtract from both sides, shown in red:

y – 7 = 3z +7 – 7

Simplifying then gives

y – 7 = 3z

Step 2: We now want to remove the 3 that is multiplying the z. To do this we divide both sides by 3 (shown in red) as this is the opposite of multiplying by 3.

Simplifying the right-hand side is simple, as the 3s cancel to leave z. The left-hand side cannot be simplified, but we can swap the two sides around to give

Note that we could also write this as

If you want to review and practice working with fractions visit the Numbers section of www.mathsupport.wordpress.com


Harder Example: Here we follow the steps of another example.

The formula to convert temperature in Celsius ( ) to temperature in Fahrenheit ( ) is

We can re-arrange this to make temperature in Celsius the subject. The aim is to get on its own, so we need to remove the variables and constants from

around . From the BIDMAS rules we know that Multiplication ( ) happens

before the Addition ( ). So when removing things we need to work in reverse order.

Step 1: We first remove the from the right-hand side. To do this we have to subtract from each side, this is shown in red and gives

Simplifying gives

Step 2: Now we want to remove the which multiplies the . There are

several ways to do this. We can multiply both sides by which is the same as

dividing both sides by . We show this multiplication of both sides in red

Notice the brackets on the left-hand side; we include these because we need to

multiply EVERYTHING on the left-hand side by

.

Simplifying gives


Notice that we have dropped the multiplication sign outside the bracket on the

left-hand side. Also that the right-hand side simplifies to just because

is the same as one. We can now swap the sides, so that is the subject,

It can help to work through algebra re-arranging in steps like this. Remember if you want to make a variable the subject of an equation, you need to remove all the variables and constants from around it using the BIDMAS rules.

6. Solving equations

To solving an equation we use the same method as re-arranging equations. The only difference is that there is just one variable, so you can re-arrange the equation to find the numerical value of this variable. This is called solving.

Example 1: Solve the following equation to find

We want to make the subject so we need to remove everything from around the on the left-hand side.

Step 1: we first want to remove the from the left-hand side. Since the opposite of subtracting is adding we to both sides.

We now simplify this to get

Which is the solution, so only needed one step. The great thing about maths is we can always check our answer. To do this we substitute into the original equation. This gives

This is correct, so we know that our answer is right.


Example 2: This is a slightly harder example. We want to solve the following equation to find

To do this we need to make the subject, so need to remove the variables, constants and operations from around the .

Step 1: add to both sides (this will remove the from the left-hand side)

Simplifying gives

Step 2: multiply both sides by (which is the opposite of dividing by )

Simplifying gives

This is the answer! As in the previous example we can check that this is correct. To do this we substitute into the original equation. This gives

We can simplify this, which gives

We can see that this is correct, so we know that is the right answer.

The key to being able to solve equations is to practicing re-arranging. So try to practice regularly. Start with the following exercises. Good Luck!


7. Now your turn

Generally the more maths you practice the easier it gets. If you make mistakes don’t worry. I generally find that if I make lots of mistakes I understand the subject better when I have finished. If you want to see videos explaining these ideas and showing the answers visit www.mathsupport.wordpress.com

A) Starting simplifying 1) 2a + 3a

2) 7a – 3a

3) 8z + 1z + 5a

4) –5a + 10a – 2z

5) – 2a – a – 8a

6) 2z –7z + 4y + 2y

7) a2 –3a2 + b – 4b

8) a2 –3a2 + a – 4a

9) b2 –3b2 + 4b – b

10) z2 –3a2 + b – 4a

B) 4a2 + 2b – 7a +3b + 8b – 9 (a – 2) + 18

1) List all the constants in the expression written above

2) List all the variables in the expression written above

3) Write the algebraic expression above in its simplest form

C) Substitution

Substitute values into the expressions below to calculate the total value of the expression: a = 7, b = -2 , y = 3, z = 1. Note you should get the same answers if you substitute the values into your answers for part A).

1) 2a + 3a

2) 7a – 3a

3) 8z + 1z + 5a

4) –5a + 10a – 2z

5) – 2a – a – 8a

6) 2z –7z + 4y + 2y

7) a2 –3a2 + b – 4b

8) a2 –3a2 + a – 4a

9) b2 –3b2 + 4b – b

10) z2 –3a2 + b – 4a


D) Practicing simplifying the following expressions using the BIDMAS rules

1) 2 (z + 2) + 8

2) 3(z – 2) – 2(z + 3)

3) 3(a – 2) – 2(b – 3)

4) 2 (z2 – 2) + 2(z – 4)

5)

6) 2 (y2 – 2) + (–2 y)2 + 4

7)

8) 2 (z2 – 2) + (2z)2 + 4

9)

10)

11) 3a(b – a2) – a2(2 – 3a)

12)

E) Re-arrange each of these equations to make the subject

1) b + 4 = a

2) 4b = y

3) 4b = y + a

4) 3b – y = 2a

5) 3b – a = 5a

6) 3(b – a) = 9a

7) 2 (b2 – 2) = 4

8) 4 = b +2y

9)

10)

11)

12)

F) Solve each of these equations to find

1) b + 2 = 8

2) 3b – 2 = 7

3) 3(b – 2) = 9

4)

5) – b + 2 = 8

6) b2 – 9 = 7

7) – 3(2 – b) = 12

8)

9) 3(b – 2) + 2b = – 9

10)

11) – 3(– b – 2) = 9

12)


G) Some word questions

1) Using T to represent the price of a cup of tea and C to represent the price of a cup of coffee write an equation to show that 2 cups of tea and 3 cups of coffee cost £5.40

a) One cup of tea costs £1.20. Substitute this value into your equation and simplify.

b) Solve this equation to find the price of one cup of coffee.

2) The formula for calculating the time taken to cook a turkey is

where T is the time in minutes needed to cook a turkey, and p is the weight in pounds.

a) How many minutes does it take to cook a 10 pound turkey

b) Re-arrange the equation to make p the subject, and then calculate how many pounds the turkey is if the cooking time is 4 hours.

2) The total cost of borrowing money can be calculated using the formula

Where is the number of years the money is borrowed for, is the interest rate as a decimal, is the amount of money borrowed and is the total amount that is repaid.

a) If the amount borrowed is £10,000, the interest rate is 5% (or 0.05 as a decimal) and the number of years is 10. Calculate the total amount that is repaid.

b) Re-arrange the equation to make the subject, and then calculate the interest rate needed if the amount borrowed is £1,000, the number of years is 3 and the total amount to be repaid is £10,000.

All answers and workings are available at www.mathsupport.wordpress.com

learning development — birkbeck, university of london

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