Equations with Brackets
An equation with brackets, such as \(8(x - 3) = 24\), treats the bracket as one quantity. In NSW Year 8 maths (NESA outcome MA4‑EQU‑C‑01) it is solved in two steps, either by dividing both sides by the number in front first or by expanding the bracket first.
You will choose the quicker method, solve equations with negative, decimal or fraction answers, handle a bracket on the right-hand side such as \(18 = 6(y + 1)\), and expand when the pronumeral also appears on the other side, as in \(3(m + 2) = 4m\). You will also find the first mistake in someone's working, and write equations for a canteen lunch order and for two shapes with equal perimeters. It is part of Year 8 Equations.
Every question with a fully worked solution.
- Equations with Brackets - Video - Equations with Brackets - Example 1 Watch
Theory
A bracket holds one quantity. In \(6(x - 4) = 42\) the whole bracket was multiplied by \(6\), so either divide both sides by \(6\) first (\(x - 4 = 7\)) or expand first (\(6x - 24 = 42\)). Both give \(x = 11\). This Year 8 maths page shows how.
Why it is two steps. In \(7(x - 2) = 35\), the \(-\,2\) happened inside the bracket first and the \(\times 7\) last. Undo the \(\times 7\) first.
Method 1: divide first. Divide both sides by the number in front, then undo the number inside. This is quicker whenever the division is tidy.
Method 2: expand first. Expanding means multiplying every term inside the bracket: \(3(x + 4) = 3x + 12\). Then collect like terms and solve. Use this when the pronumeral is on both sides.
The bracket may be on the right. \(45 = 9(k - 2)\) gives \(5 = k - 2\), so \(k = 7\). Write the answer with the pronumeral on the left.
The number in front may be negative. Divide both sides by it, sign and all; or expand with the sign: \(-2(4 - x) = -8 + 2x\).
Brackets on both sides. Expand both, take the smaller pronumeral term from both sides, then undo the number: \(2(x + 1) = 3(x - 4)\) gives \(2x + 2 = 3x - 12\), so \(x = 14\).
Modelling. When a group is repeated, put the group in a bracket: \(4\) bags each holding \(m\) apples and \(2\) pears hold \(4(m + 2)\) pieces of fruit.
Two routes to the same answer:
| Equation | Steps (both sides) | Solution |
|---|---|---|
| \(4(x - 1) = 20\) | \(\div 4\), then \(+\,1\) | \(x = 6\) |
| \(-3(x + 2) = 9\) | \(\div (-3)\), then \(-\,2\) | \(x = -5\) |
| \(24 = 8(y - 1)\) | \(\div 8\), then \(+\,1\) | \(y = 4\) |
| \(5(x + 3) = 6x\) | expand to \(5x + 15 = 6x\), then \(-\,5x\) | \(x = 15\) |
Divide first
- Divide both sides by the number in front of the bracket.
- Undo the number inside the bracket.
- Check in the original equation.
Expand first
- Expand every bracket: multiply each term inside.
- Collect the pronumeral terms on one side (take away the smaller one).
- Solve the two-step equation that is left and check.
| \(3 \times x + 3 \times 4\) | \(=\) | \(15\) |
| \(3x + 12\) | \(=\) | \(15\) |
| \(3x + 12 - 12\) | \(=\) | \(15 - 12\) |
| \(3x\) | \(=\) | \(3\) |
| \(\dfrac{3x}{3}\) | \(=\) | \(\dfrac{3}{3}\) |
| \(x\) | \(=\) | \(1\) |
\(x = 1\). Dividing first is quicker here: \(x + 4 = 5\), so \(x = 1\).
| \(\dfrac{6(x - 4)}{6}\) | \(=\) | \(\dfrac{42}{6}\) |
| \(x - 4\) | \(=\) | \(7\) |
| \(x - 4 + 4\) | \(=\) | \(7 + 4\) |
| \(x\) | \(=\) | \(11\) |
\(x = 11\). Check: \(6(11 - 4) = 6 \times 7 = 42\).
(i) \(-4(x + 1) = 10\)
(ii) \(5(p - 2) = 4\)
| \(x + 1\) | \(=\) | \(\dfrac{10}{-4} = -2.5\) |
| \(x\) | \(=\) | \(-2.5 - 1 = -3.5\) |
| \(p - 2\) | \(=\) | \(\dfrac{4}{5}\) |
| \(p\) | \(=\) | \(\dfrac{4}{5} + 2 = 2\dfrac{4}{5}\) |
(i) \(x = -3.5\). Check: \(-4(-3.5 + 1) = -4 \times (-2.5) = 10\).
(ii) \(p = 2\dfrac{4}{5}\). Check: \(5 \times \dfrac{4}{5} = 4\).
(i) \(2(x + 6) = 5x\)
(ii) \(2(4 - x) = 5 - 7x\)
(i) Expand to \(2x + 12 = 5x\), then subtract \(2x\) from both sides.
(ii) Expand to \(8 - 2x = 5 - 7x\). Add \(7x\) to both sides, then subtract \(8\).
| \(12\) | \(=\) | \(3x\) |
| \(x\) | \(=\) | \(4\) |
| \(8 + 5x\) | \(=\) | \(5\) |
| \(5x\) | \(=\) | \(-3\) |
| \(x\) | \(=\) | \(-\dfrac{3}{5}\) |
(i) \(x = 4\). Check: \(2(4 + 6) = 20 = 5 \times 4\).
(ii) \(x = -\dfrac{3}{5}\). Check: both sides equal \(9\dfrac{1}{5}\).
| \(2x + 2\) | \(=\) | \(3x - 12\) |
| \(2x + 2 - 2x\) | \(=\) | \(3x - 12 - 2x\) |
| \(2\) | \(=\) | \(x - 12\) |
| \(14\) | \(=\) | \(x\) |
\(x = 14\). Check: \(2(14 + 1) = 30\) and \(3(14 - 4) = 30\).
(ii) Noor picks a number, adds \(4\), then multiplies the result by \(3\). She gets \(-6\). What was her number?
(iii) An equilateral triangle has sides of \((x + 5)\) cm and a square has sides of \(x\) cm. Their perimeters are equal. Find \(x\).
| \(5(21 - d)\) | \(=\) | \(80\) |
| \(21 - d\) | \(=\) | \(16\) |
| \(d\) | \(=\) | \(5\) |
| \(3(n + 4)\) | \(=\) | \(-6\) |
| \(n + 4\) | \(=\) | \(-2\) |
| \(n\) | \(=\) | \(-6\) |
| \(3(x + 5)\) | \(=\) | \(4x\) |
| \(3x + 15\) | \(=\) | \(4x\) |
| \(x\) | \(=\) | \(15\) |
(i) Each friend gets \(\$5\) off, paying \(\$16\).
(ii) Her number was \(-6\). Check: \((-6 + 4) \times 3 = -6\).
(iii) \(x = 15\): both perimeters are \(60\) cm.
Common pitfalls
Frequently asked questions
How do you solve an equation with brackets?
Divide both sides by the number in front of the bracket, or expand the bracket first, then solve the equation that is left.
What does expand mean?
Multiply every term inside the bracket by the number outside: \(3(x + 4) = 3x + 12\).
When should I expand instead of dividing?
When the pronumeral is on both sides, or when the division would not come out tidily.
What if the number in front is negative?
Divide both sides by that negative number, or expand carefully with the sign: \(-2(x - 3) = -2x + 6\).
How do I solve brackets on both sides?
Expand both brackets, take the smaller pronumeral term from both sides, then finish as a two-step equation.