10 Ways to Support your Child with Algebra

Whilst algebra is commonly introduced to children in year 6, algebra style questions are often included in 11+ examinations. These questions range in difficulty so it is good to be aware of the different types of questions that your child may come up against.

1) Write an Expression

This is when you might be asked to create an expression from a word sentence.

E.g. Write an expression for ‘There are c cars in a car park, then 5 more cars arrive’.

Answer:

c + 5

2) Combine “like” terms

This is when you are asked to collect the ‘like’ terms together so that you can then simplify the equation or expression.

E.g. Simplify 3r + 3b – 2r + 2b – b

Answer:

Combine ‘r’s and then ‘b’s

r + 3b + 2b – b
r + 4b

3) Expand brackets

If you are required to expand brackets, make sure you expand the brackets carefully.

E.g. Expand the following brackets

1)     3(a + b) = 3a + 3b

2)     4(y + 6) = 4y + 24

3)     2(2a + 3b – 4c) = 4a + 6b – 8c

And a bit more challenging:

4)     (x + 4) (x + 3) =

Multiply everything in the first bracket by the second bracket:

= x2 + 3x + 4x + 12
= x2 + 7x + 12

4) Using BODMAS/BIDMAS

When working out a calculation, don’t forget to go in the order of BODMAS (or BIDMAS):

Brackets; Orders; Division and Multiplication; Addition and Subtraction.

E.g. 7 + (6 x 52 + 3)
7 + (6 x 25 + 3)
7 + (150 + 3)
7 + (153)
7 + 153
= 160

5) Write and Solve an Equation

This is when you create an equation from a word problem and then solve it.

E.g. I’m thinking of a number. When I subtract the number from 45, I get the same answer as when I double the number. What’s my number?

Method:

45 – n = 2n
45 = 3n
15 = n

E.g 2. Jim runs a fruit stall at a local market. He sells oranges in packets of 3 and apples in packets of 4.

George buys 5 packets of oranges and some packets of apples. In total, he has 31 pieces of fruit. How many packets of apples did George buy?

Method:

Let’s call packs of oranges ‘x’ and apples ‘y’

3x + 4y = 31
15 + 4y = 31
4y = 16
y = 4

6) Using Simultaneous Equations to solve a Word Problem

This is when you have two variables and you have the total number for both.

E.g. Mike has 27 vehicles to service at his garage. Some are cars (4 wheels) and the rest are motorcycles (2 wheels). Altogether, the 27 vehicles have 78 wheels. How many cars does Mike have to service?

Method:

Let’s call the cars ‘c’ and the motorbikes ‘m’

c + m = 27

4c + 2m = 78

Now multiply the simpler equation by 4 so that you can remove the ’c’s.

4c + 4m = 108

Now line them up with the larger numbers on the top.

4c + 4m = 108
4c + 2m = 78

Now subtract each one downwards.

2m = 30
m = 15

As c + m = 27, c = 12.

7) Using the Grid Method to solve a Word Problem

This is when you have 2 or 3 variables and usually a total cost for each one.

E.g. Three apples and a banana cost 32p. 6 apples and a banana cost 53p. How much does one banana cost?

Method:

Apples	Bananas	Total
3	1	32
6	1	53

 

Now subtract the largest from smallest*

Apples	Bananas	Total
6	1	53
3	1	32
3	0	21

If 3 apples are 21p, 1 apple is 7p and so one banana is 11p.

*Note that you do not always subtract one from the other – you need to think what you could do to get the answer you require (i.e. sometimes you add).

NB: Sometimes you can use either the grid method OR simultaneous equations! Try this question using simultaneous equations.

8) Substituting and Eliminating

These are often the hardest questions! These questions are when you create an equation for each piece of information. You then substitute and eliminate terms, so that you are left with one term, allowing you to find the answer.

E.g. Three children, Rana, Sarah and Tina, are collecting leaves. Rana collects twice as many as Sarah. Sarah collects one and a half times as many as Tina. Between them they collect 198 leaves. How many leaves did each child collect?

Method:

r + s + t = 198

r = 2s

s = 1.5t

2s + 1.5t + t = 198 (replace the r and s)
3t + 1.5t + t = 198 (replace the s)
5.5t = 198
11t = 396 (double it to make it easier to divide)
t = 36

s = 1.5 x 36
s = 54

r = 2s
r = 108

9) Draw a Chart and use Trial and Error

If algebra proves to be too complicated, trial and error can often be the fastest option.

E.g. Three children, Rana, Sarah and Tina, are collecting leaves. Rana collects twice as many as Sarah. Sarah collects one and a half times as many as Tina. Between them they collect 198 leaves. How many leaves did each child collect?

Method:

Let’s start with Tina. Choose a number under 198 but choose an even number so that Sarah’s number will be a whole number. 

Rana	Sarah	Tina	Total	198?
150	75	50	275	High
120	60	40	220	High
90	45	30	165	Low
108	54	36	198	Perfect!

 

10) Draw a Picture to represent the Problem

It may also be possible to draw the word problem as a picture, which may prove easier than algebra. This is particularly the case for questions involving races!

E.g. In the same time that Tom can run 100 metres, Jerry can only run 60 metres. Tom and Jerry have a competition. When Jerry has run 100 metres, Tom sets off in pursuit. How many metres will Tom run before he catches Jerry?

Answer: 250 metres.