Year 7 Mathematics | Victorian Curriculum 2.0
Linear equations
Topic 06 | Number & Algebra | Practice
How this topic is organised

The Year 7 core is solving one-variable linear equations with natural-number solutions (whole positive numbers), and checking solutions by substitution (VC2M7A03).

Extension covers equations whose solutions are negative or fractional, and equations with a pronumeral on both sides. These appear formally in Year 8 (VC2M8A01 family).

Year 7 core

By the end of this topic you should be able to:

  • solve one-step linear equations using inverse operations,
  • solve two-step linear equations by undoing operations in reverse order,
  • verify a solution by substituting it back in,
  • translate a simple worded problem into an equation and solve it.
Why does 'balance both sides' work?

An equation is a statement that two things are equal — like a pair of balanced scales. The equals sign is a promise: “left side and right side have the same value.” If you add 333 to one side but not the other, the promise breaks and the equation becomes false. That is why every operation you perform must happen to both sides — you are keeping the truth intact while uncovering the unknown.

Where you'll see this
  • Shopping puzzles: “three pens and a ruler cost $8, pens are $2 each — what’s the ruler?”
  • Age problems: “Ava is 3 years older than her sister; in 5 years their ages sum to 33.”
  • Travel: “a taxi charges $4 flag-fall plus $2/km; a trip cost $29 — how far?”
  • Target setting: “I want to save $500; I save $25/week — how many weeks?”
Worked example 0 Real-world example: saving for a goal

You want to buy a $500 gaming console. You already have $80 saved and add $25 each week. How many weeks until you can buy it?

  1. Write the equation: 80+25w=50080 + 25w = 50080+25w=500.
  2. Subtract 808080: 25w=42025w = 42025w=420.
  3. Divide by 252525: w=16.8w = 16.8w=16.8.
  4. You can’t save a fraction of a week, so you need 171717 weeks.
  5. Check: 80+25×17=80+425=50580 + 25 \times 17 = 80 + 425 = 50580+25×17=80+425=505. ✓ (enough with $5 spare).

Key idea: the equation translates the English sentence into maths; solving it means undoing the operations to find the unknown.

1. The balance idea

An equation is a balanced pair of scales: whatever is on the left equals whatever is on the right. If you change one side, you must do the exact same thing to the other side to keep the scales level.

x+5=12⟹x+5−5=12−5⟹x=7.x + 5 = 12 \quad\Longrightarrow\quad x + 5 - 5 = 12 - 5 \quad\Longrightarrow\quad x = 7.x+5=12⟹x+5−5=12−5⟹x=7.
Golden rule

Whatever you do to one side, do to the other. Otherwise the scales tip and the equation is no longer true.

2. Inverse operations

To “undo” an operation, apply its inverse.

Inverse operations

Addition and subtraction are inverses

x+a=b  ⇒  x=b−ax + a = b \;\Rightarrow\; x = b - ax+a=b⇒x=b−a. Example: x+7=12⇒x=5x + 7 = 12 \Rightarrow x = 5x+7=12⇒x=5.

Multiplication and division are inverses

ax=b  ⇒  x=baax = b \;\Rightarrow\; x = \dfrac{b}{a}ax=b⇒x=ab​. Example: 3x=18⇒x=63x = 18 \Rightarrow x = 63x=18⇒x=6.

3. One-step equations

Worked example 1 One-step addition/subtraction

Solve   x+9=23\;x + 9 = 23x+9=23.

Subtract 999 from both sides: x=23−9=14x = 23 - 9 = 14x=23−9=14.

Check: 14+9=2314 + 9 = 2314+9=23.

Worked example 2 One-step multiplication/division

Solve   5x=45\;5x = 455x=45.

Divide both sides by 555: x=45÷5=9x = 45 \div 5 = 9x=45÷5=9.

Check: 5×9=455 \times 9 = 455×9=45.

4. Two-step equations

When there are two operations, undo them in reverse order: deal with addition/subtraction first, then with multiplication/division.

Worked example 3 Two-step equation

Solve   3x+4=19\;3x + 4 = 193x+4=19.

  1. Subtract 444 from both sides: 3x=153x = 153x=15.
  2. Divide both sides by 333: x=5x = 5x=5.

Check: 3×5+4=15+4=193 \times 5 + 4 = 15 + 4 = 193×5+4=15+4=19.

Worked example 4 Variable in a fraction

Solve   x4+3=8\;\dfrac{x}{4} + 3 = 84x​+3=8.

  1. Subtract 333 from both sides: x4=5\dfrac{x}{4} = 54x​=5.
  2. Multiply both sides by 444: x=20x = 20x=20.

Check: 20÷4+3=5+3=820 \div 4 + 3 = 5 + 3 = 820÷4+3=5+3=8.

5. From words to equations

Identify the unknown, give it a letter, write down what the sentence says.

Worked example 5 A number puzzle

“I think of a number, double it, add 555, and the result is 232323. What is the number?”

  1. Let the number be nnn.
  2. Write the equation: 2n+5=232n + 5 = 232n+5=23.
  3. Solve: 2n=182n = 182n=18, so n=9n = 9n=9.

Check: 2×9+5=232 \times 9 + 5 = 232×9+5=23.


Practice: Year 7 core

All answers are natural numbers (positive whole numbers).

Fluency

One-step equations

    1. Solve   x+8=20\;x + 8 = 20x+8=20.
    2. Solve   y−6=13\;y - 6 = 13y−6=13.
    3. Solve   4x=28\;4x = 284x=28.
    4. Solve   y3=7\;\dfrac{y}{3} = 73y​=7.
    5. Solve   x+17=30\;x + 17 = 30x+17=30.
    6. Solve   a−9=1\;a - 9 = 1a−9=1.
    7. Solve   7m=42\;7m = 427m=42.
    8. Solve   x6=4\;\dfrac{x}{6} = 46x​=4.
Fluency

Two-step equations

    1. Solve   2x+3=11\;2x + 3 = 112x+3=11.
    2. Solve   3y−5=7\;3y - 5 = 73y−5=7.
    3. Solve   5x+1=16\;5x + 1 = 165x+1=16.
    4. Solve   4x−3=9\;4x - 3 = 94x−3=9.
    5. Solve   7a−4=24\;7a - 4 = 247a−4=24.
    6. Solve   x2+1=6\;\dfrac{x}{2} + 1 = 62x​+1=6.
    7. Solve   m3−4=1\;\dfrac{m}{3} - 4 = 13m​−4=1.
    8. Solve   x+52=4\;\dfrac{x + 5}{2} = 42x+5​=4.
    9. Solve   2x+13=5\;\dfrac{2x + 1}{3} = 532x+1​=5.
    10. Solve   3(x+2)=18\;3(x + 2) = 183(x+2)=18.
Reasoning

Verify and reason

    1. Verify, without solving, whether x=6x = 6x=6 is a solution of   2x+4=16\;2x + 4 = 162x+4=16.
    2. Verify whether y=3y = 3y=3 is a solution of   5y−1=15\;5y - 1 = 155y−1=15.
    3. Zara solves 2x+6=142x + 6 = 142x+6=14 by writing ”2x=14+6=202x = 14 + 6 = 202x=14+6=20, so x=10x = 10x=10”. Explain her mistake and give the correct answer.
    4. Find the missing number: solve   □+7=15\;\square + 7 = 15□+7=15 and check by substitution.
    5. Write an equation of your own whose solution is x=8x = 8x=8, and verify it by substitution.
Problem solving

Worded problems

    1. Four times a number, increased by 777, is 313131. Find the number.
    2. A pencil costs $2 and a ruler costs $r. Three pencils and a ruler cost $8 in total. Find rrr.
    3. Ava is xxx years old. Her sister is 333 years older. In 555 years, the sum of their ages will be 333333. How old is Ava now?
    4. A taxi charges a $4 flag-fall and $2.50 per kilometre. A trip cost $29. How long was the trip?
    5. The perimeter of an isosceles triangle with two equal sides of length xxx cm and base 888 cm is 303030 cm. Find xxx.
    6. Five consecutive natural numbers sum to 454545. Find the smallest of them. (Hint: call it nnn and write n+(n+1)+…n + (n+1) + \ldotsn+(n+1)+…)

Extension

Beyond Year 7 core

The questions below have negative or fractional solutions, or a pronumeral on both sides. These skills appear formally in Year 8 and rely on confidence with negative numbers.

Practice: Extension

Reasoning

Non-positive and two-sided equations

    1. Solve   x+11=5\;x + 11 = 5x+11=5.
    2. Solve   −2m=14\;-2m = 14−2m=14.
    3. Solve   2x+9=3\;2x + 9 = 32x+9=3.
    4. Solve   −3x+5=20\;-3x + 5 = 20−3x+5=20.
    5. Solve   10−2x=4\;10 - 2x = 410−2x=4.
    6. Solve   5x−2=2x+10\;5x - 2 = 2x + 105x−2=2x+10.
    7. Solve   4y+7=y+19\;4y + 7 = y + 194y+7=y+19.
    8. Solve   x2+x3=10\;\dfrac{x}{2} + \dfrac{x}{3} = 102x​+3x​=10. (Hint: multiply both sides by 666.)
    9. The sum of three consecutive integers is −9-9−9. Find them.
    10. A number increased by 40%40\%40% gives 848484. What was the original number?
Year 7 Mathematics study companion | Practice