Question 1
Rearrange the following equations to make x the subject.
For all questions, give your answer in its simplest form.
1(a) 3x = 8
ANSWER: Multiple Choice (Type 1)
A: x = 24
B: x = \dfrac{3}{8}
C: x = 5
D: x = \dfrac{8}{3}
Answer: D
Marks = 1
1(b) 5x = 2y
ANSWER: Mutiple Choice (Type 1)
A: x = \dfrac{5}{2y}
B: x = 10y
C: x = \dfrac{2}{5}y
D: x = \dfrac{5}{2}y
Answer: C
Marks = 1
1(c) \dfrac{5}{x} = 2y
ANSWER: Mutiple Choice (Type 1)
A: x = \dfrac{5}{2y}
B: x = 10y
C: x = \dfrac{2}{5}y
D: x = \dfrac{5}{2}y
Answer: A
Marks = 1
1(d) 10 - x = 5
ANSWER: Mutiple Choice (Type 1)
A: x = 5
B: x = 15
C: x = 2
D: x = -5
Answer: A
Marks = 1
Question 2
2(a) Rearrange b - 2x = n to make b the subject.
ANSWER: Multiple Choice (Type 1)
A: b = n- 2x
B: b = n + 2x
C: b = \dfrac{n}{2x}
D: b = -\dfrac{n}{2x}
Answer: B
Marks = 1
2(b) Rearrange t^2 = 3x + 1 to make x the subject.
ANSWER: Multiple Choice (Type 1)
A: x = \dfrac{1 - t^2}{3}
B: x = \dfrac{t^2}{3} - 1
C: x = \dfrac{t^2}{3} + 1
D: x = \dfrac{t^2 -1}{3}
Answer: D
Workings:
t^2 - 1 = 3x
x = \dfrac{t^2 -1}{3}
Marks = 1
2(c) Rearrange 3p + 4 = 6 - q to make q the subject.
Simplify your answer.
ANSWER: Multiple Choice (Type 1)
A: q = 3p + 10
B: q = -3p - 2
C: q = 2 - 3p
D: q = \dfrac{6}{3p + 4}
Answer: C
Workings:
q = 6 - (3p + 4) = 6 -3p - 4
q = 2 - 3p
Marks = 1
2(d) Given \sqrt{x-2} = 3a find x in terms of a
ANSWER: Multiple Choice (Type 1)
A: x = 2 + 9a^2
B: x = 2 + 3a^2
C: x = 2 - 9a^2
D: x = 2 - 3a^2
Answer: A
Workings:
x - 2 = (3a)^2 = 9a^2
x = 2 + 9a^2
Marks = 2
Question 3
3(a) Make S the subject of 2S + 3 = T - 5
ANSWER: Multiple Choice (Type 1)
A: S = \dfrac{T-2}{2}
B:S = \dfrac{T-5}{2} +3
C: S = \dfrac{T-8}{2}
D:S = \dfrac{T-5}{2} -3
Answer: C
Workings:
2S = T - 8
S = \dfrac{T-8}{2}
Marks = 2
3(b) Make y the subject of \dfrac{3 + x}{y} = 2x + 1
ANSWER: Multiple Choice (Type 1)
A: y = \dfrac{3 + x}{2x + 1}
B: y = 2 - x
C: y = \dfrac{3 + x}{2x} - 1
D: y = \dfrac{3 + x}{2x - 1}
Answer: A
Workings:
3 + x = y(2x + 1)
y = \dfrac{3 + x}{2x + 1}
Marks = 2
Question 4
4(a) Rearrange a^2 + b^2 = c^2 to make b the subject
ANSWER:
b = \sqrt{c^2 - a^2}
b = \sqrt{c^2 + a^2}
b = c + a
b = \sqrt{c + a}
Workings:
Step 1: b^2 = c^2 -a^2
Step 2: b = \sqrt{c^2 - a^2}
Marks = 2
4(b) Rearrange \sqrt{a+b} = c to make b the subject
ANSWER:
b = c^2 -a
b = \sqrt{c^2 + a^2}
b = (c + a)^2
b = \sqrt{c }+a
Workings:
Step 1: a+b = c^2
Step 2: b = c^2 -a
Marks = 2
Question 5
Rearrange the following to make x the subject of the equation
\dfrac{x+1}{2} +10 = y
ANSWER:
x = 2y-21
x = \dfrac{2}{y-21}
x = 2y-19
x = \dfrac{1}{y-19}
Workings:
Step 1: \dfrac{x+1}{2} +10 = y
Step 2:\dfrac{x+1}{2} = y-10
Step 3: x+1 = 2(y-10)
Step 4: x+1 = 2y-20
Step 5: x = 2y-20-1
Step 6: x = 2y-21
Question 6
Patrick is trying to rearrange the following formula in order to make x the subject of the formal
\dfrac{x+y}{3} + z = \dfrac{3}{y}
He performs the following steps:
\begin{aligned}\text{Line } 1:\,\,\, \dfrac{x+y}{3} + z &= \dfrac{3}{y} \\\\ \text{Line } 2:\,\,\,\,\,\,\,\,\,\,\,\,\, \dfrac{x+y}{3} &= \dfrac{3}{y-z} \\\\ \text{Line } 3:\,\,\,\,\,\,\,\,\,\,\,\,\,\, x+y &= 3\bigg(\dfrac{3}{y-z} \bigg) \\\\ \text{Line } 4:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,x &= 3\bigg(\dfrac{3}{y-z}\bigg) -y \end{aligned}
Patrick has made a mistake with his rearrangement.
6(a) Give the correct equation for \text{Line } 2
\dfrac{x+y}{3} = \dfrac{3}{y} - z
\dfrac{x+y}{3} = \dfrac{3}{y+z}
\dfrac{x+y}{3} = \dfrac{3}{y} + z
\dfrac{x+y}{3} = \dfrac{3}{y z}
Working
\dfrac{x+y}{3} = \dfrac{3}{y} - z
we must subtract z from both sides, placing it outside the fraction.
6(b) Give the correct equation for \text{Line } 3
x+y = 3\bigg(\dfrac{3}{y} - z\bigg)
x+y = 3\bigg(\dfrac{3}{y-z}\bigg)
x+y = \bigg(\dfrac{3}{y-z}\bigg)^3
x+y = 3\bigg(\dfrac{3}{y} - z\bigg)^3
Working
x+y = 3\bigg(\dfrac{3}{y} - z\bigg)
We must multiply both sides by three giving this result.
6(c) Give the correct equation for \text{Line } 4
x = 3\bigg(\dfrac{3}{y} - z\bigg) -y
x = 3\bigg(\dfrac{3}{y} - z\bigg)^3 - y
x = \bigg(\dfrac{3}{y-z}\bigg)^3 - y
x = 3\bigg(\dfrac{3}{y} - z\bigg) \times y
Answer:
\begin{aligned}\text{Line } 1:\,\,\, \dfrac{x+y}{3} + z &= \dfrac{3}{y} \\\\ \text{Line } 2:\,\,\,\,\,\,\,\,\,\,\,\,\, \dfrac{x+y}{3} &= \dfrac{3}{y} - z \\\\ \text{Line } 3:\,\,\,\,\,\,\,\,\,\,\,\,\,\, x+y &= 3\bigg(\dfrac{3}{y} - z\bigg) \\\\ \text{Line } 4:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,x &= 3\bigg(\dfrac{3}{y} - z\bigg) -y \end{aligned}For \text{Line } 4 we simply subtract y from both sides