KS3 Maths · Year 7 · Chapter 15: Equations and formulae · Lesson 1

Equations with brackets

Year 7About 60 minutesNo calculator56 marks

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Date

Do Now

Quick questions to start.

A1
Needed today
Expand $4 (2 x - 5)$ .
[1 mark]
Answer
A2
Needed today
Solve $3 x + 7 = 34$ .
[1 mark]
Answer
A3
Last lesson
A circle has radius $10$ cm. Work out its area. Give your answer in terms of $π$ .
[1 mark]
Answer
A4
From chapter 13
$5$ pens cost £ $3.50$ . Work out the cost of $8$ pens.
[1 mark]
Answer

Example, then your turn

Year 7

Your teacher works through each example with you. Then try the one beside it.

Example 1

Solve $6 (x - 3) = 42$ . Use two different methods.

Hint Method 1: divide both sides by

6

first. Method 2: expand the bracket first.

B1
Solve $4 (x - 9) = 28$ .
[2 marks]
Answer

Example 2

Solve $5 (11 - x) = 20$ .

Hint Divide by

5

. Then add

x

to both sides so that the

x

term is positive.

B2
Solve $4 (13 - x) = 30$ .
[2 marks]
Answer

Example 3

Solve $4 (2 x + 7) = 6$ . Give your answer as a mixed number.

Hint Expand the bracket. The answer will be negative, and it is not a whole number.

B3
Solve $2 (3 x + 10) = 7$ . Give your answer as a mixed number.
[3 marks]
Answer

Example 4

Solve $\frac{2}{3} (x - 4) = 10$ .

Hint Multiply both sides by

3

first, so there is no fraction.

B4
Solve $\frac{3}{4} (x + 2) = 15$ .
[3 marks]
Answer

Example 5

The perimeter of this rectangle is $46$ cm. Write an equation with a bracket. Solve it to find $x$ and the length of the rectangle.

Not drawn accurately

Hint Perimeter

= 2 \times (length + width)

B5
The triangle is equilateral. Its perimeter is $51$ cm. Write an equation with a bracket. Solve it to find $x$ and the length of one side.
[3 marks]
Not drawn accurately
Answer $x = \dots\dots$ side $\dots\dots$ cm

Practice

Year 7

C1
Work out the value of $\frac{3}{4} (x + 6)$ when $x = 14$ .
[1 mark]
Answer
C2
Solve $54 = 6 (x + 5)$ .
[2 marks]
Answer

C3
Solve $5 x + 32 = 7$ .
[2 marks]
Answer
C4
Solve $2 (9 - 4 x) = 50$ .
[3 marks]
Answer
C5
Solve $3 (x - 1.4) = 7.9$ . Give your answer correct to $2$ decimal places.
[3 marks]
Answer
C6
Solve $\frac{1}{2} (4 x - 7) = 9$ . Give your answer as a mixed number.
[3 marks]
Answer

Exam-style questions

GCSE Foundation

D1
Ben solves $3 (x + 4) = 27$ . He writes: $3 x + 4 = 27$ , so $3 x = 23$ and $x = 7 \frac{2}{3}$ .
Ben's answer is wrong. Explain what Ben did wrong, and solve the equation correctly.
[3 marks]
Answer

D2
Solve each equation. You must show your working.
(a)
$2 (5 x + 3) = 41$
[3 marks]
Answer
(b)
$\frac{x - 7}{4} = - 3$
[2 marks]
Answer
Total for D2: 5 marks
D3
The diagram shows an isosceles triangle. The perimeter of the triangle is $49$ cm.
Not drawn accurately
(a)
Show that $7 x = 49$ .
[2 marks]
(b)
Work out the length of the base of the triangle.
[2 marks]
Answer
Total for D3: 4 marks
D4
A school buys $6$ boxes of pencils. Each box holds $x$ pencils. The art teacher takes $4$ pencils out of each box. There are now $150$ pencils left in the boxes.
(a)
Write down an equation in $x$ . Use a bracket.
[2 marks]
Answer
(b)
Solve your equation to find the number of pencils that were in each box.
[2 marks]
Answer
(c)
How many pencils were in the $6$ boxes before the art teacher took any?
[1 mark]
Answer
Total for D4: 5 marks
D5
Here are four equations.
$A 3 (x - 2) = 15 B 2 (10 - x) = 6 C \frac{x + 5}{3} = 4 D 4 (2 x - 9) = 24$
Three of the equations have the same solution. Which equation is the odd one out? You must show your working.
[4 marks]
AnswerEquation $\dots\dots$

Extension

Stretch

No route is given. The hints are at the end of the sheet. Use one at a time.

E1
The mean of the three expressions $2 (x + 3)$ , $3 (x - 1)$ and $x + 12$ is $21$ . Work out the value of $x$ .
[4 marks]
Hint 1 · What to try
The mean is $21$ , so the three expressions add up to $3 \times 21$ .
Hint 2 · The first line
$2 (x + 3) + 3 (x - 1) + (x + 12) = 63$
Hint 3 · The full method
Expand: $2 x + 6 + 3 x - 3 + x + 12 = 6 x + 15$ . So $6 x + 15 = 63$ , $6 x = 48$ and $x = 8$ .
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can solve an equation with a bracket by dividing first or by expanding first.
I can solve equations whose answers are negative numbers or fractions.
I can clear a fraction from an equation by multiplying.
I can write and solve an equation from a perimeter or a word problem.

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