E2.5 – Equations

E2.5 Test — Equations
15 Questions  —  Section A: Recall  |  Section B: Application  |  Section C: Challenge
Instructions:
  • Answer all 15 questions in the spaces provided.
  • Show all your working — marks are given for the method, not just the final answer.
  • Section A — Recall: short answers and one or two steps.
  • Section B — Application: full worked solutions.
  • Section C — Challenge: multi-part questions. Answer each part (a), (b), (c) separately.
  • Give exact answers in surd form where asked — do not round.
Section A — Recall
Questions 1–5  |  Definitions and short answers
1.

State the difference between an expression and an equation.

2.

Solve $2x + 7 = 15$.

3.

Write an expression for the product of two consecutive even numbers.

4.

Write down the standard form of a quadratic equation.

5.

Make $x$ the subject of the formula $y = 2x + 1$.

Section B — Application
Questions 6–10  |  Worked solutions
6.

Solve $4(x – 1) = 2x + 6$.

7.

Solve the fractional equation $\dfrac{x}{3x – 2} = 2$.

8.

Solve the simultaneous equations:

$3x + y = 11$
$2x – y = 4$

9.

Solve $x^2 + 7x + 12 = 0$ by factorisation.

10.

Make $x$ the subject of the formula $y = \sqrt{x – 2}$.

Section C — Challenge
Questions 11–15  |  Extended and multi-part answers
11.
A cinema sells adult tickets for $x$ dollars and child tickets for $y$ dollars.
• 3 adult tickets and 2 child tickets cost \$23.
• 1 adult ticket and 2 child tickets cost \$13.

(a) Write two equations to represent this information.

(b) Solve your equations to find the price of an adult ticket and a child ticket.

12.

Solve $\dfrac{3}{x + 1} + \dfrac{2}{x – 1} = 1$.

13.

(a) Write $x^2 + 8x + 10$ in completed square form.

(b) Hence solve $x^2 + 8x + 10 = 0$, giving your answers in surd form.

(c) Solve $2x^2 – 4x – 3 = 0$ using the quadratic formula, giving your answers in surd form.

14.

Consider the simultaneous equations $y = x + 1$ and $y = x^2 – 1$.

(a) Solve the equations to find all pairs of values of $x$ and $y$.

(b) Explain why this pair of equations has two solution pairs.

15.

(a) Make $x$ the subject of $y = \dfrac{x + 4}{x – 2}$.

(b) Make $r$ the subject of $V = \pi r^2 h$.

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