AQA Level 2 Further Maths · Level 2 Certificate · Chapter 2: Algebra II · Lesson 2

Rearranging formulae

About 70 minutesNo calculator61 marks

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Date

Do Now

The first questions are what today's lesson needs. The others bring back earlier topics.

A1
Needed today
Make $x$ the subject of $y = \frac{x}{4} - 7$ .
[2 marks]
Answer
A2
Last lesson
Factorise $15 x^{2} + 7 x - 2$ .
[2 marks]
Answer
A3
From chapter 1
Work out the coefficient of $x^{2}$ in the expansion of $(1 + 2 x)^{5}$ .
[2 marks]
Answer
A4
From chapter 1
Simplify $\frac{12}{3} + 27$ .
[2 marks]
Answer

Example, then your turn

Level 2

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

Example 1

Make $v$ the subject of $E = \frac{1}{2} m v^{2}$ , where $v > 0$ .

Hint Undo the steps in reverse order: the half and the

m

first, the square last.

B1
Make $r$ the subject of $V = \frac{1}{3} π r^{2} h$ , where $r > 0$ .
[2 marks]
Answer

Example 2

Make $l$ the subject of $T = 2 π \frac{l}{g}$ .

Hint Get the square root on its own first. Then square both sides.

B2
Make $k$ the subject of $m = 4 - 2 k + 1$ .
[3 marks]
Answer

Example 3

Make $x$ the subject of $a (x - 3) = b (x + 2)$ .

Hint Expand both brackets. Collect every term in

x

on one side and everything else on the other. Then factorise out

x

B3
Make $y$ the subject of $3 (y - p) = p y + 4$ .
[3 marks]
Answer

Example 4

Make $x$ the subject of $y = \frac{2 x + 3}{x - 4}$ .

Hint Multiply both sides by

x - 4

to clear the fraction. Then it is the same as the last example.

B4
Make $t$ the subject of $w = \frac{5 - 3 t}{2 t + 1}$ .
[3 marks]
Answer

Example 5

Make $v$ the subject of $\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$ .

Hint Get

\frac{1}{v}

on its own, write the other side as one fraction, then turn both sides upside down.

B5
Make $b$ the subject of $\frac{2}{a} - \frac{3}{b} = c$ .
[3 marks]
Answer

Practice

Level 2

C1
Make $u$ the subject of $s = \frac{( u + v ) t}{2}$ .
[2 marks]
Answer
C2
Make $x$ the subject of $y = 5 x^{3} - 2$ .
[2 marks]
Answer

C3
Make $x$ the subject of $x^{2} + a^{2} = (x + b)^{2}$ .
[3 marks]
Answer
C4
Make $a$ the subject of $\frac{a}{a + 1} = k$ .
[3 marks]
Answer
C5
Make $x$ the subject of $y = \frac{x}{3} + \frac{x}{5}$ .
[2 marks]
Answer
C6
Make $d$ the subject of $F = \frac{9 q}{d ^{2}}$ , where $d > 0$ .
[2 marks]
Answer

Exam-style questions

AQA exam style

D1
Ali makes $x$ the subject of $y = x^{2} + 9$ . He writes " $y = x + 3$ , so $x = y - 3$ ".
Explain the mistake Ali makes. Make $x$ the subject of $y = x^{2} + 9$ correctly.
[3 marks]
Answer

D2
Make $k$ the subject of $m = \frac{2 k ^{3} + 5}{k ^{3} - 4}$ .
[5 marks]
Answer
D3
$y = x^{2} + 10 x + 31$ , where $x > - 5$ .
(a)
Show that $x^{2} + 10 x + 31 = (x + 5)^{2} + 6$ .
[1 mark]
(b)
Hence make $x$ the subject of $y = x^{2} + 10 x + 31$ .
[2 marks]
Answer
(c)
Work out the value of $x$ when $y = 87$ .
[1 mark]
Answer
Total for D3: 4 marks
D4
Two pipes fill a tank. Pipe A alone takes $p$ hours. Pipe B alone takes $q$ hours. With both pipes on, the tank takes $T$ hours to fill, where $\frac{1}{T} = \frac{1}{p} + \frac{1}{q}$ .
(a)
Make $q$ the subject of the formula.
[3 marks]
Answer
(b)
Pipe A alone takes $6$ hours. Both pipes together take $4$ hours. How long does pipe B take alone?
[1 mark]
Answer $\dots\dots\dots$ hours
Total for D4: 4 marks
D5
The shape is a square of side $x$ cm with a semicircle on its top side. The dashed side is not part of the outline.
Not drawn accurately
(a)
Show that the perimeter, $P$ cm, of the shape is given by $P = 3 x + \frac{1}{2} π x$ .
[2 marks]
(b)
Make $x$ the subject of $P = 3 x + \frac{1}{2} π x$ . Give your answer as a single fraction.
[2 marks]
Answer
Total for D5: 4 marks

Extension

Stretch

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

E1
Make $x$ the subject of $y = \frac{x ^{2} + 1}{x ^{2} - 1}$ . Hence find the values that $y$ can never take.
[5 marks]
Hint 1 · What to try
Treat $x^{2}$ as the subject first, the same way as $y = \frac{2 x + 3}{x - 4}$ . Then think about when a square root can be taken.
Hint 2 · The first line
$y (x^{2} - 1) = x^{2} + 1$ , so $x^{2} (y - 1) = y + 1$ and $x^{2} = \frac{y + 1}{y - 1}$ .
Hint 3 · The full method
So $x = \pm \frac{y + 1}{y - 1}$ . This needs $\frac{y + 1}{y - 1} \geq 0$ , which fails when $- 1 < y < 1$ ; and $y = 1$ makes the bottom zero. So $y$ can never take values with $- 1 < y \leq 1$ .

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can change the subject of a formula with powers and roots in it.
I can rearrange a formula where the subject appears more than once.
I can rearrange formulae with fractions and reciprocals.
I can rearrange a formula and then use it to work out a value.

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