Year 7 Equations: free practice test with answers

Sample equations questions with answers

Try each one before you open the answer.

1. Introduction to equations

   Is $x = 4$ a solution of $3x + 5 = 17$?

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   Key idea: A value is a solution if substituting it makes both sides equal.

   1. Left side: $3 \times 4 + 5 = 12 + 5 = 17$
   2. Right side: $17$
   3. Both sides are equal, so $x = 4$ is a solution

   Answer: Yes

2. Solving equations by inspection

   Solve by inspection:

   (a) $x + 9 = 23$ (b) $7m = 63$ (c) $\dfrac{k}{4} = 6$ (d) $20 - p = 8$

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   Key idea: Ask: what number makes the equation true?

   1. (a) What plus $9$ gives $23$? $14$, since $14 + 9 = 23$
   2. (b) $7$ times what is $63$? $9$, since $7 \times 9 = 63$
   3. (c) What divided by $4$ is $6$? $24$, since $24 \div 4 = 6$
   4. (d) $20$ minus what is $8$? $12$, since $20 - 12 = 8$

   Answer: (a) $x = 14$ (b) $m = 9$ (c) $k = 24$ (d) $p = 12$

3. Equivalent equations

   Write the equation that results when you add $5$ to both sides of $x - 5 = 12$.

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   Key idea: Doing the same operation to both sides gives an equivalent equation.

   1. Left: $x - 5 + 5 = x$
   2. Right: $12 + 5 = 17$
   3. New equation: $x = 17$

   Answer: $x = 17$

4. Solving equations algebraically

   Solve $2x + 5 = 17$, then verify your solution by substitution.

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   Key idea: Undo the operations in reverse order, subtract first, then divide.

   1. Subtract $5$ from both sides:\

   $$2x + 5 - 5 = 17 - 5 \implies 2x = 12$$

   2. Divide both sides by $2$:\

   $$x = \frac{12}{2} = 6$$

   3. Verify by substituting $x = 6$ back into the original equation:\

   $$2(6) + 5 = 12 + 5 = 17 \checkmark$$

   Answer: $x = 6$

5. Equations with fractions

   Solve:

   (a) $\dfrac{x}{5} + 3 = 7$ (b) $\dfrac{2m - 4}{3} = 6$ (c) $\dfrac{3k}{4} = 12$

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   Key idea: Remove the fraction by multiplying both sides by the denominator at the right moment.

   1. (a) Subtract $3$: $\dfrac{x}{5} = 4$; multiply by $5$: $x = 20$
   2. (b) Multiply by $3$: $2m - 4 = 18$; add $4$: $2m = 22$; divide by $2$: $m = 11$
   3. (c) Multiply by $4$: $3k = 48$; divide by $3$: $k = 16$

   Answer: (a) $x = 20$ (b) $m = 11$ (c) $k = 16$

6. Equations with brackets

   Solve $3(x + 2) = 21$.

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   Key idea: Expand the brackets first, then solve the two-step equation.

   1. Expand the brackets:\

   $$3(x + 2) = 3x + 6$$

   So the equation becomes $3x + 6 = 21$. 2. Subtract $6$ from both sides:\

   $$3x = 21 - 6 = 15$$

   3. Divide both sides by $3$:\

   $$x = \frac{15}{3} = 5$$

   4. Verify by substituting $x = 5$ into the original equation:\

   $$3(5 + 2) = 3 \times 7 = 21 \checkmark$$

   Answer: $x = 5$

7. Using formulas

   The perimeter of a rectangle is given by the formula $P = 2l + 2w$. Find $P$ when $l = 9$ and $w = 4$.

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   Key idea: Replace each pronumeral with its given value, then evaluate.

   1. Write the formula: $P = 2l + 2w$
   2. Substitute $l = 9$ and $w = 4$: $P = 2 \times 9 + 2 \times 4$
   3. Multiply: $P = 18 + 8$
   4. Add: $P = 26$

   Answer: $P = 26$

8. Using equations to solve problems

   Aisha is saving up to buy a book that costs $29. She already has $11 saved and earns $6 per hour doing odd jobs. How many hours does Aisha need to work to have exactly enough money to buy the book?

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   Key idea: Translate the situation into a linear equation, solve it, then verify.

   1. Let $h$ be the number of hours Aisha needs to work. Her total savings will be $6h + 11$ dollars. Set this equal to $29$:\

   $$6h + 11 = 29$$

   2. Subtract $11$ from both sides:\

   $$6h = 29 - 11 = 18$$

   3. Divide both sides by $6$:\

   $$h = \frac{18}{6} = 3$$

   4. Verify by substituting $h = 3$:\

   $$6(3) + 11 = 18 + 11 = 29 \checkmark$$

   Answer: Aisha needs to work $3$ hours.

Free worksheets for this chapter

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

• Year 7 linear equations worksheet
• Year 7 algebra worksheet