Year 8 Equations and inequalities: free practice test with answers

Sample equations and inequalities questions with answers

Try each one before you open the answer.

1. Reviewing equations

   Solve $3x + 7 = 22$.

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   Key idea: Use inverse operations to isolate $x$, undoing addition before division.

   1. Subtract $7$ from both sides: $3x + 7 - 7 = 22 - 7 \implies 3x = 15$
   2. Divide both sides by $3$: $x = \frac{15}{3} = 5$
   3. Check by substituting back: $3(5) + 7 = 15 + 7 = 22$ ✓

   Answer: $x = 5$

2. Equivalent equations

   Solve $5x - 8 = 17$ and check your answer.

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   Key idea: Form equivalent equations by doing the same operation to both sides.

   1. Add $8$ to both sides: $5x = 25$
   2. Divide both sides by $5$: $x = 5$
   3. Check: $5 \times 5 - 8 = 17$ ✓

   Answer: $x = 5$

3. Equations with fractions

   Solve for $x$: $\dfrac{x + 3}{4} = 5$.

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   Key idea: Undo each operation in reverse order to isolate the pronumeral.

   1. Multiply both sides by 4: $\dfrac{x+3}{4} \times 4 = 5 \times 4$, giving $x + 3 = 20$.
   2. Subtract 3 from both sides: $x + 3 - 3 = 20 - 3$.
   3. Simplify: $x = 17$.
   4. Check by substituting back: $\dfrac{17 + 3}{4} = \dfrac{20}{4} = 5$ ✓

   Answer: $x = 17$

4. Equations with the pronumeral on both sides

   Solve $7x - 4 = 3x + 20$.

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   Key idea: Collect the pronumeral terms on one side by subtracting the smaller one from both sides.

   1. Subtract $3x$ from both sides: $4x - 4 = 20$
   2. Add $4$: $4x = 24$, so $x = 6$
   3. Check: $7 \times 6 - 4 = 38$ and $3 \times 6 + 20 = 38$ ✓

   Answer: $x = 6$

5. Solving simple quadratic equations

   Solve $x^2 = 49$.

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   Key idea: A positive number has two square roots, one positive and one negative.

   1. $x = \pm\sqrt{49}$
   2. $x = 7$ or $x = -7$, since $7^2 = 49$ and $(-7)^2 = 49$

   Answer: $x = 7$ or $x = -7$

6. Formulas and relationships

   A plumber charges a call-out fee of $80 plus $60 for each hour worked. Write a linear equation for the total cost $C$ (in dollars) in terms of the number of hours worked $h$.

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   Key idea: A linear equation has the form $C = mh + b$, where $m$ is the rate of change (cost per hour) and $b$ is the fixed starting value (call-out fee).

   1. Identify the fixed cost (intercept): the call-out fee is $80, so $b = 80$.
   2. Identify the rate of change (slope): the cost increases by $60 per hour, so $m = 60$.
   3. Write the equation:\

   $$C = 60h + 80$$

   4. Check: for $h = 1$, $C = 60(1) + 80 = 140$. One hour should cost $80 + $60 = $140. ✓

   Answer: $C = 60h + 80$

7. Applications of equations

   A plumber charges a $60 call-out fee plus $45 for each hour worked. Mia pays a total of $195. Write and solve a linear equation to find how many hours the plumber worked.

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   Key idea: Translate the word problem into a linear equation, then solve using inverse operations.

   1. Let $h$ be the number of hours worked. The total cost is: $60 + 45h = 195$
   2. Subtract $60$ from both sides: $45h = 135$
   3. Divide both sides by $45$: $h = 3$
   4. Check: $60 + 45 \times 3 = 60 + 135 = 195$ ✓

   Answer: The plumber worked for $3$ hours.

8. Inequalities

   Write an inequality for each statement.

   (a) $x$ is greater than $-3$.

   (b) $x$ is at most $12$.

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   Key idea: "At most" means less than or equal to, so the endpoint is included.

   1. (a) Greater than: $x > -3$
   2. (b) At most $12$ means $12$ or less: $x \le 12$

   Answer: (a) $x > -3$ (b) $x \le 12$

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
• Year 8 algebra worksheet
• Year 8 speed, distance and time worksheet
• Year 8 profit, loss and discounts worksheet