2.2 Changing the Subject

Quick Notes & Formulae

Essential concepts for rearranging formulae and solving equations.

Rearranging Formulae

Use inverse operations: addition ↔ subtraction, multiplication ↔ division. Work in reverse order of BIDMAS. Whatever you do to one side, do to the other.

With Brackets

Either divide both sides by the coefficient first, or expand then rearrange:

y = 3 (x + 2) \Rightarrow \frac{y}{3} = x + 2

y = 3 x + 6

With Fractions

Multiply both sides by the denominator first to eliminate fractions:

y = \frac{x}{4} + 5 \Rightarrow y - 5 = \frac{x}{4} \Rightarrow 4 (y - 5) = x

Cross-Multiplication

For

\frac{a}{b} = \frac{c}{d}

: multiply diagonally to get

a d = b c

. Useful when variable is in denominator.

Solving Linear Equations

Collect variable terms on one side, constants on the other. Use inverse operations to isolate the variable. Check by substitution.

Reverse Problems

Given the solution, substitute back to find unknown constants. Form an equation and solve for the missing value.

Past Paper Style Questions

Question 1: Simple Rearranging

Make

x

the subject.

(a)

y = x + 7

[1]

(b)

a = 5 x

[1]

(c)

p = \frac{x}{4}

[1]

(d)

m = 3 x - 2

[2]

(e) Verify your answer in (d) by substituting

x = 4

. [1]

Question 2: Rearranging with Brackets

Make

x

the subject.

(a)

y = 4 (x + 3)

[2]

(b)

a = 2 (3 x - 5)

[2]

(c)

p = 5 - 3 (x - 2)

[2]

(d)

k = 7 (2 x + 1) - 4

[2]

(e) State the first step you perform in part (b). [1]

Question 3: Rearranging with Fractions

Make

x

the subject.

(a)

y = \frac{x}{3} + 5

[2]

(b)

a = \frac{2 x - 1}{4}

[2]

(c)

m = \frac{5 - x}{2}

[2]

(d)

p = \frac{3 x + 7}{6}

[2]

(e) Check your answer in (a) by substituting

x = 9

. [1]

Question 4: Formula from Geometry (Exam Style)

The perimeter of a rectangle is

P = 2 l + 2 w

(a) Make

l

the subject. [2]

(b) Make

w

the subject. [1]

l

when

P = 50

and

w = 8

. [2]

(d) Find

w

when

P = 64

and

l = 12

. [2]

(e) Explain why you must divide by 2 first when rearranging. [1]

Question 5: Solving Linear Equations

Solve the following.

(a)

5 x + 3 = 28

[2]

(b)

7 x - 4 = 3 x + 12

[2]

(c)

4 (x - 2) = 3 x + 6

[2]

(d)

6 - 2 x = 4 x + 18

[2]

(e) Check your answer in (c). [1]

Question 6: Reverse – Find Missing Constant

The equation

3 x + k = 19

has solution

x = 5

(a) Substitute

x = 5

into the equation. [1]

(b) Find

k

. [1]

(d) Solve your equation to verify

x = 5

. [2]

(e) Explain why substitution works here. [1]

Question 7: Rearranging Multi-Step Formula

Make

x

the subject.

(a)

y = 3 x + 2 a

[2]

(b)

k = 5 - 2 x + 4 a

[2]

(c)

m = 7 x - 3 a

[2]

(d)

p = 4 (2 x - a)

[2]

(e) Identify which parts involve collecting like terms. [1]

Question 8: Context Formula (Speed)

The formula for speed is

v = \frac{d}{t}

(a) Make

d

the subject. [1]

(b) Make

t

the subject. [1]

t

when

v = 60

and

d = 180

. [2]

(d) Find

d

when

v = 25

and

t = 3.2

. [2]

(e) State why cross-multiplication is used. [1]

Question 9: Reverse Equation Solving

The equation

a x + 7 = 3 x + 19

has solution

x = 4

(a) Substitute

x = 4

into the equation. [1]

(b) Form an equation in

a

. [1]

a

. [2]

(d) Write the completed equation. [1]

(e) Solve it again to confirm

x = 4

. [2]

Question 10: Challenge – Combined Rearranging + Solving

The formula

y = \frac{3 x - 5}{2}

(a) Make

x

the subject. [3]

(b) Find

x

when

y = 7

. [1]

\frac{3 x - 5}{2} = 10

. [2]

(d) Verify your answer in (c) by substitution. [1]

(e) Explain the importance of removing fractions first when solving. [1]

Question 11: Rearranging Linear Formulae

Make

x

the subject.

(a)

y = 5 x - 9

[1]

(b)

p = 3 - 4 x

[2]

(c)

m = \frac{x - 7}{3}

[2]

(d)

k = \frac{2 x + 5}{4}

[2]

(e) Check your answer to (d) by substituting

x = 3

into the original formula. [1]

Question 12: Rearranging with Brackets and Collecting Terms

Make

x

the subject.

(a)

y = 2 (x - 5) + 7

[2]

(b)

a = 3 (2 x + 1) - 5

[2]

(c)

p = 4 - 2 (x + 3)

[2]

(d)

t = 6 (x - 2) - 3 (x + 4)

[3]

(e) In part (d), state at which step you collect like terms. [1]

Question 13: Rearranging Fractional Equations

Make

x

the subject.

(a)

y = \frac{x}{5} - 2

[2]

(b)

a = \frac{3 x - 4}{7}

[2]

(c)

m = \frac{2 - x}{4}

[2]

(d)

p = \frac{5 x + 1}{3} + 2

[2]

(e) Check your answer in (b) by substituting

x = 6

. [1]

Question 14: Solve Linear Equations (One Variable)

Solve.

(a)

9 x - 5 = 31

[2]

(b)

4 x + 7 = 3 x + 19

[2]

(c)

6 (2 x - 3) = 3 (3 x + 4)

[2]

(d)

\frac{x - 5}{3} = 4

[2]

(e) For (c), show one clear expansion step before solving. [1]

Question 15: Reverse – Find the Missing Constant

The equation

5 x - k = 13

has solution

x = 4

(a) Substitute

x = 4

into the equation. [1]

(b) Find

k

. [1]

(d) Solve your completed equation to confirm

x = 4

. [2]

(e) Explain why the solution is unique. [1]

Question 16: Rearranging a Geometry Formula

The area of a trapezium is

A = \frac{1}{2} (a + b) h

(a) Make

h

the subject. [2]

(b) Make

a

the subject. [3]

h

when

A = 84

a = 10

b = 18

. [2]

(d) Find

a

when

A = 60

b = 14

h = 6

. [2]

(e) State one reason why brackets must be used when rearranging for

a

. [1]

Question 17: Rearranging with Negative Coefficients

Make

x

the subject.

(a)

y = 7 - 3 x

[2]

(b)

p = - 2 x + 9

[2]

(c)

m = 4 - \frac{x}{2}

[2]

(d)

k = 5 - \frac{2 x - 1}{3}

[3]

(e) For part (a), explain why dividing by −3 changes the sign. [1]

Question 18: Speed/Time/Distance (Changing Subject + Solving)

The formula is

v = \frac{d}{t}

(a) Make

t

the subject. [1]

(b) A car travels 270 km at 90 km/h. Find

t

. [2]

d

. [2]

(d) A cyclist travels 18 km in 1.2 h. Find

v

. [2]

(e) Convert your speed in (d) to m/s (3 s.f.). [2]

Question 19: Reverse – Find Parameter So Equation Has Given Solution

The equation

(a - 2) x + 5 = 3 x + 17

has solution

x = 6

(a) Substitute

x = 6

into the equation. [1]

(b) Form an equation in

a

. [1]

a

. [2]

(d) Write the completed equation. [1]

(e) Solve the completed equation to confirm

x = 6

. [2]

Question 20: Challenge – Rearrange Then Solve

The formula is

y = \frac{4 x - 3}{x + 2}

(where

x \neq - 2

(a) Multiply both sides by

(x + 2)

and expand. [2]

(b) Make

x

the subject in terms of

y

. [3]

x

when

y = 2

. [1]

(d) Find

x

when

y = - 1

. [1]

(e) State any values of

y

that would make the rearranged formula invalid, and explain. [2]