Year 8 Linear relationships: free practice test with answers

Sample linear relationships questions with answers

Try each one before you open the answer.

1. The Cartesian plane

   In which quadrant does each point lie?

   (a) $(3, -5)$

   (b) $(-2, -7)$

   (c) $(-4, 6)$

   Show the answer

   Key idea: The quadrants are numbered anticlockwise from the top right: first $(+, +)$, second $(-, +)$, third $(-, -)$, fourth $(+, -)$.

   1. (a) $x$ positive and $y$ negative: fourth quadrant
   2. (b) $x$ negative and $y$ negative: third quadrant
   3. (c) $x$ negative and $y$ positive: second quadrant

   Answer: (a) Fourth quadrant (b) Third quadrant (c) Second quadrant

2. Using rules, tables and graphs to explore linear relationships

   Complete the table of values for $y = 5 - 3x$. Give the $y$-values in order.

   $x$	$-2$	$-1$	$0$	$1$	$2$
   $y$

   Show the answer

   Key idea: Substitute each $x$-value into the rule.

   1. $x = -2$: $5 - 3(-2) = 11$; $x = -1$: $5 + 3 = 8$; $x = 0$: $5$.
   2. $x = 1$: $5 - 3 = 2$; $x = 2$: $5 - 6 = -1$.

   Answer: 11, 8, 5, 2, −1

3. Finding the rule using a table of values

   The table shows some points on a straight line.

   $x$	$0$	$1$	$2$	$3$
   $y$	$2$	$4$	$6$	$8$

   (a) Write the rule for $y$ in terms of $x$.

   (b) Find $y$ when $x = 10$.

   (c) Find $x$ when $y = 30$.

   Show the answer

   Answer: (a) $y = 2x + 2$ (b) $y = 22$ (c) $x = 14$ ($2x + 2 = 30$)

4. Using graphs to solve linear equations

   (a) Copy and complete the table of values for $y = 3x - 2$.

   $x$	$-1$	$0$	$1$	$2$	$3$
   $y$

   (b) Use the table (or the graph of the line) to solve $3x - 2 = 4$.

   Show the answer

   Key idea: The solution of $3x - 2 = 4$ is the $x$-value of the point on the line $y = 3x - 2$ where $y = 4$.

   1. (a) $3(-1) - 2 = -5$, $3(0) - 2 = -2$, $3(1) - 2 = 1$, $3(2) - 2 = 4$, $3(3) - 2 = 7$
   2. (b) $y = 4$ when $x = 2$, so $x = 2$
   3. Check: $3 \times 2 - 2 = 4$ ✓

   Answer: (a) −5, −2, 1, 4, 7 (b) $x = 2$

5. The x- and y-intercepts

   Find the $x$- and $y$-intercepts of the line $y = 4x - 12$.

   Show the answer

   Key idea: The $y$-intercept is where $x = 0$; the $x$-intercept is where $y = 0$.

   1. $y$-intercept: $y = 4(0) - 12 = -12$, so $(0, -12)$
   2. $x$-intercept: $0 = 4x - 12$, so $x = 3$, giving $(3, 0)$

   Answer: $x$-intercept $(3, 0)$, $y$-intercept $(0, -12)$

6. Gradient

   Find the gradient of the line passing through $(2, 5)$ and $(6, 17)$.

   Show the answer

   Key idea: Gradient $= \dfrac{\text{rise}}{\text{run}} = \dfrac{y_2 - y_1}{x_2 - x_1}$.

   1. Rise: $17 - 5 = 12$; run: $6 - 2 = 4$
   2. Gradient $= \dfrac{12}{4} = 3$

   Answer: $3$

7. Gradient–intercept form

   Write the equation $2y - 6x = 8$ in the form $y = mx + c$. State the gradient and $y$-intercept, then find the coordinates of the point where the line crosses the $x$-axis.

   Show the answer

   Key idea: Rearrange into $y = mx + c$ to read off the gradient $m$ and $y$-intercept $c$; set $y = 0$ to find the $x$-intercept.

   1. Add $6x$ to both sides: $2y = 6x + 8$
   2. Divide everything by $2$: $y = 3x + 4$
   3. So the gradient is $m = 3$ and the $y$-intercept is $c = 4$, giving the point $(0, 4)$.
   4. Set $y = 0$ to find the $x$-intercept: $0 = 3x + 4 \implies 3x = -4 \implies x = -\frac{4}{3}$
   5. Check: $3\left(-\frac{4}{3}\right) + 4 = -4 + 4 = 0$ ✓

   Answer: Gradient $= 3$, $y$-intercept $= 4$, $x$-intercept at $\left(-\dfrac{4}{3},\, 0\right)$

8. Applications of straight line graphs

   A school is hiring a bus for an excursion. Company A charges a flat fee of $320 plus $4 per student. Company B charges a flat fee of $200 plus $6 per student. Write a linear equation for the total cost of each company in terms of the number of students $n$, find the number of students at which both companies charge the same amount, and state which company is cheaper when 80 students attend.

   Show the answer

   Key idea: Write an equation for each option, set them equal to find the break-even point, then compare for the given number of students.

   1. Write the equations:\

   $$A = 4n + 320 B = 6n + 200$$

   2. Set $A = B$ to find the break-even point:\

   $$4n + 320 = 6n + 200$$

   3. Subtract $4n$ from both sides:\

   $$320 = 2n + 200$$

   4. Subtract 200, then divide by 2:\

   $$\begin{aligned} 120 &= 2n \\ n &= 60 \end{aligned}$$

   5. For $n = 80$: $A = 4(80) + 320 = 640$; $B = 6(80) + 200 = 680$. Since $640 < 680$, Company A is cheaper.
   6. Check break-even: $A = 4(60) + 320 = 560$; $B = 6(60) + 200 = 560$. ✓

   Answer: Both companies charge the same for 60 students. For 80 students, Company A is cheaper at $640.

Free worksheets for this chapter

Worked examples and a free PDF with answers for each topic.

• Year 8 linear equations and graphs worksheet
• Year 8 linear relations in real-life problems worksheet