KS3 Maths · Year 7 · Chapter 15: Equations and formulae · Lesson 3

More complex equations

Year 7About 70 minutesNo calculator56 marks

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Date

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Quick questions to start.

A1
Needed today
Write down both numbers whose square is $64$ .
[1 mark]
Answer
A2
Last lesson
Solve $3 (x + 2) = x + 14$ .
[1 mark]
Answer
A3
From chapter 12
Work out $\frac{2}{3}$ of $45$ .
[1 mark]
Answer
A4
From chapter 4
Increase £ $80$ by $15%$ .
[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 $3 (t + 7) = 2 (t - 4)$ . Check your answer.

Hint Expand both brackets, then take

2 t

from both sides.

B1
Solve $4 (y + 6) = 3 (y - 2)$ .
[3 marks]
Answer

Example 2

Solve $\frac{3}{5} (x + 4) = x - 6$ .

Hint Multiply both sides by

5

. The right-hand side becomes

5 (x - 6)

B2
Solve $\frac{2}{7} (x + 9) = x - 3$ . Give your answer as a mixed number.
[3 marks]
Answer

Example 3

Solve $\frac{1}{3} (x - 2) = \frac{1}{4} (x + 5)$ .

Hint Multiply both sides by

12

, a multiple of

3

and of

4

B3
Solve $\frac{x + 7}{2} = \frac{x + 22}{5}$ .
[3 marks]
Answer

Example 4

Solve $3 t^{2} - 7 = 68$ .

Hint Get

t^{2}

on its own. A number has two square roots: one positive and one negative.

B4
Solve $6 n^{2} - 11 = 205$ .
[3 marks]
Answer

Example 5

Lena is $n$ years old. Four years ago, her age was two-thirds of what her age will be in six years' time. Write an equation and solve it to find Lena's age.

Hint Four years ago she was

n - 4

. In six years she will be

n + 6

B5
Kai is $k$ years old. Three years ago, his age was half of what his age will be in nine years' time. Write an equation and solve it to find Kai's age.
[3 marks]
Answer

Practice

Year 7

C1
Solve $5 (10 - y) = 2 (13 - y)$ .
[3 marks]
Answer
C2
Solve $\frac{r + 11}{3} = r - 5$ .
[2 marks]
Answer

C3
Solve $22 = 3 k^{2} - 5$ .
[2 marks]
Answer
C4
Write down all the solutions of each equation. If there are none, write "none".
(i) $x^{2} - 81 = 0$
(ii) $9 x^{2} = 0$
(iii) $x^{2} + 16 = 0$
[3 marks]
Answer(i) $\dots\dots$ (ii) $\dots\dots$ (iii) $\dots\dots$
C5
Explain why the equation $4 (x - 3) = 4 x + 5$ has no solution.
[2 marks]
C6
Solve $\frac{x ^{2}}{2} - 3 = 47$ .
[2 marks]
Answer

Exam-style questions

GCSE Foundation

D1
Zara solves $3 x^{2} = 147$ . She writes: $x^{2} = 49$ , so $x = 7$ .
Zara's answer is not complete. Explain why, and give the full answer.
[2 marks]
Answer

D2
Line $P$ is split into $4$ equal parts. Each part is $(x - 6)$ cm long. Line $Q$ is split into $2$ equal parts. Each part is $(x + 5)$ cm long. The two lines are the same length.
Not drawn accurately
(a)
Write down an equation in $x$ .
[1 mark]
Answer
(b)
Work out the length of each line.
[3 marks]
Answer
Total for D2: 4 marks
D3
Rectangle $A$ and rectangle $B$ have the same area. Work out this area. You must show your working.
[5 marks]
Not drawn accurately
Answer
D4
The square and the equilateral triangle have the same perimeter. Work out this perimeter.
[4 marks]
Not drawn accurately
Answer
D5
The area of the square is $25$ cm² more than the area of the rectangle.
Not drawn accurately
(a)
Show that $x^{2} = 25$ .
[2 marks]
(b)
The equation $x^{2} = 25$ has two solutions. Explain why only one of them is the value of $x$ here.
[1 mark]
(c)
Work out the perimeter of the rectangle.
[2 marks]
Answer
Total for D5: 5 marks

Extension

Stretch

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

E1
Solve $2 (x^{2} + 5) = 3 x^{2} - 6$ .
[3 marks]
Hint 1 · What to try
Expand the bracket, then collect the $x^{2}$ terms on one side, as you would with $x$ terms.
Hint 2 · The first line
$2 x^{2} + 10 = 3 x^{2} - 6$
Hint 3 · The full method
Take $2 x^{2}$ from both sides: $10 = x^{2} - 6$ , so $x^{2} = 16$ and $x = 4$ or $x = - 4$ .
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can solve equations with brackets on both sides.
I can clear fractions from an equation by multiplying.
I can solve an equation with a square and give both solutions.
I can write an equation from a shape or an age problem and solve it.

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