Year 9 Maths Advanced Equations

Equations with Brackets

Q: What happens when there is a negative outside the bracket?

Every term inside changes sign. For example -3(x - 4) = -3x + 12, because -3 times -4 is +12.

20 practice questions 2 video lessons Theory + worked examples

Equations with brackets are solved by expanding first, collecting like terms, then solving. This NSW Year 9 Advanced maths subtopic takes extra care with a negative in front of a bracket (NESA outcome MA5‑EQU‑C‑01).

You will solve equations such as $5(2 - 3x) = -8 - 6x$ and $9 - 7(3x + 5) = x - 2(x + 4)$, including nested brackets and fractions inside brackets. You will use equations to find the sides of an isosceles triangle and the prices of pies and drinks, and show that one equation is true for every value of $x$, not just one. It is part of Year 9 Advanced Equations.

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Practice questions

Every question with a fully worked solution.

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Watch 2 video(s)

Equations with Brackets - Video - Solving an equation with distributive property on both sides Watch
Equations with Brackets - Video - Solve a Linear Equation with Parentheses and Variables on Both Sides: -a(x+b)+c=dx+e Watch

Theory

To solve an equation with brackets, expand the brackets using the distributive law, collect like terms on each side, then solve as usual. This Year 9 Advanced maths page covers a negative number or a minus sign in front of a bracket, brackets on both sides, brackets inside brackets, fractions and problems about perimeters and costs.

Grouping symbols (brackets) show that everything inside is treated as one quantity. To remove them, use the distributive law: the number outside multiplies every term inside, $a(b + c) = ab + ac$.

A negative number outside changes the sign of every term inside: $-4(x - 2) = -4x + 8$. A minus sign on its own means $-1$: $-(3 - x) = -3 + x$.

After expanding, collect like terms on each side separately, so each side has at most one $x$ term and one number. Then solve as an equation with the pronumeral on one or both sides.

With nested brackets such as $3[x - 2(x + 1)]$, work from the inside out: expand and simplify the inner bracket first.

Sometimes it is quicker not to expand. In $5(x - 3) = 20$ you can divide both sides by $5$ first: $x - 3 = 4$.

The number outside multiplies every term inside: $3(x + 4) = 3x + 12$.

A negative number outside changes the sign of every term inside: $-2(x - 5) = -2x + 10$.

The distributive law, and the sign rules it uses:

\[a(b + c) = ab + ac \qquad -a(b - c) = -ab + ac \qquad -(b - c) = -b + c\]

a (b + c) = a b + a c

Expression	Expanded
$4(2x - 3)$	$8x - 12$
$-3(x + 6)$	$-3x - 18$
$-5(2 - x)$	$-10 + 5x$
$-(4x - 1)$	$-4x + 1$
$6\left(\dfrac{x}{3} - \dfrac{1}{2}\right)$	$2x - 3$

Fractions and brackets together. In \(\dfrac{2(x - 1)}{3} = x - 2\), multiply both sides by \(3\) first and put brackets round the right side: \(2(x - 1) = 3(x - 2)\). Then expand. For fractions in front of brackets, such as \(\dfrac{1}{2}(x + 4) - \dfrac{1}{3}(x - 3) = 2\), multiply every term by the lowest common multiple, \(6\), first.

How to solve an equation with brackets

Clear fractions first, if there are any, by multiplying every term by the lowest common multiple of the denominators.
Expand every bracket, inner brackets first. Multiply every term inside, and watch the signs.
Collect like terms on each side.
Solve: get the $x$ terms on one side and the numbers on the other, then divide.
Check by substituting into the original equation, brackets and all.

Worded problems

Define the pronumeral, then write the other quantities in terms of it (a cap costs $(t - 8)$ dollars if it is $\$8$ less than a T-shirt at $t$ dollars).
Use brackets whenever a number multiplies a whole expression: $5$ caps cost $5(t - 8)$, not $5t - 8$.
Equal lengths or totals give the equation: equal sides of an isosceles triangle, equal perimeters, or a total cost.
Identity check: if both sides expand to the same expression, every value of $x$ is a solution; test two values to confirm.

Example 1 — A minus sign in front

Solve $4(3 - 2x) - (x - 5) = 8$.

Solution

Expand both brackets. The minus sign in front of $(x - 5)$ means $-1 \times (x - 5) = -x + 5$.

$12 - 8x - x + 5$	$=$	$8$
$17 - 9x$	$=$	$8$
$-9x$	$=$	$-9$
$x$	$=$	$\mathbf{1}$

17 - 9 x = 8, x = 1

Check: $4(3 - 2) - (1 - 5) = 4 + 4 = 8$.

Example 2 — Brackets on both sides

Solve $3(2x + 5) = 7(x - 1) + 2$.

Solution

Expand and simplify each side, then collect the $x$ terms on the side with more $x$.

$6x + 15$	$=$	$7x - 7 + 2$
$6x + 15$	$=$	$7x - 5$
$20$	$=$	$x$

6 x + 15 = 7 x - 5, x = 20

So $x = \mathbf{20}$. Check: LHS $= 3(45) = 135$, RHS $= 7(19) + 2 = 135$.

Example 3 — Brackets inside brackets

Solve $2[5 - 3(x - 1)] = x + 4$.

Solution

Work from the inside out. Simplify the inner bracket first.

$5 - 3(x - 1)$

$=$

$5 - 3x + 3 = 8 - 3x$

$2(8 - 3x)$	$=$	$x + 4$
$16 - 6x$	$=$	$x + 4$
$12$	$=$	$7x$
$x$	$=$	$\mathbf{\dfrac{12}{7}}$

16 - 6 x = x + 4, x = \frac{12}{7}

Leave the answer as a fraction. Check: $x - 1 = \dfrac{5}{7}$, so LHS $= 2\left(5 - \dfrac{15}{7}\right) = \dfrac{40}{7}$ and RHS $= \dfrac{12}{7} + 4 = \dfrac{40}{7}$.

Example 4 — Equal perimeters

An equilateral triangle has sides of $(2x + 3)$ cm. A square has sides of $(x + 4)$ cm. The two shapes have the same perimeter. Find $x$ and the perimeter.

Solution

Each perimeter is a number of equal sides times the side length.

$3(2x + 3)$	$=$	$4(x + 4)$
$6x + 9$	$=$	$4x + 16$
$2x$	$=$	$7$
$x$	$=$	$\mathbf{3.5}$

3 (2 x + 3) = 4 (x + 4), x = 3.5

Perimeter $= 3(2 \times 3.5 + 3) = 3 \times 10 = \mathbf{30}$ cm. Check with the square: $4(3.5 + 4) = 30$ cm.

Common pitfalls

Multiply every term inside. $3(2x - 4)$ is $6x - 12$, not $6x - 4$.

Negative times negative. $-4(3 - x) = -12 + 4x$. The second term becomes positive.

The lone minus sign. $10 - (x + 2)$ is $10 - x - 2 = 8 - x$. The minus applies to both terms in the bracket.

Inside first. With nested brackets, simplify the inner bracket before expanding the outer one.

Keep brackets when multiplying a side. Multiplying $x - 1$ by $4$ gives $4(x - 1)$, not $4x - 1$.

Frequently asked questions

How do you solve equations with brackets?

Expand the brackets by multiplying the number outside by every term inside, collect like terms on each side, then solve the linear equation. Check by substituting into the original equation.

What happens when there is a negative outside the bracket?

Every term inside changes sign. For example $-3(x - 4) = -3x + 12$, because $-3 \times (-4) = +12$.

Should I expand the brackets or divide first?

If the whole side is a single bracket times a number, such as $4(x + 2) = 28$, dividing both sides by $4$ first is quicker. Otherwise expand.

How do you deal with brackets inside brackets?

Work from the inside out. Expand and simplify the innermost bracket, then expand the next one out, collecting like terms as you go.

What if both sides expand to the same expression?

Then the equation is an identity: it is true for every value of $x$. Subtracting the $x$ terms leaves a true statement such as $-5 = -5$.

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Using Formulae

Expression	Expanded
\(4(2x - 3)\)	\(8x - 12\)
\(-3(x + 6)\)	\(-3x - 18\)
\(-5(2 - x)\)	\(-10 + 5x\)
\(-(4x - 1)\)	\(-4x + 1\)
\(6\left(\dfrac{x}{3} - \dfrac{1}{2}\right)\)	\(2x - 3\)

\(12 - 8x - x + 5\)	\(=\)	\(8\)
\(17 - 9x\)	\(=\)	\(8\)
\(-9x\)	\(=\)	\(-9\)
\(x\)	\(=\)	\(\mathbf{1}\)

\(6x + 15\)	\(=\)	\(7x - 7 + 2\)
\(6x + 15\)	\(=\)	\(7x - 5\)
\(20\)	\(=\)	\(x\)

\(2(8 - 3x)\)	\(=\)	\(x + 4\)
\(16 - 6x\)	\(=\)	\(x + 4\)
\(12\)	\(=\)	\(7x\)
\(x\)	\(=\)	\(\mathbf{\dfrac{12}{7}}\)

\(3(2x + 3)\)	\(=\)	\(4(x + 4)\)
\(6x + 9\)	\(=\)	\(4x + 16\)
\(2x\)	\(=\)	\(7\)
\(x\)	\(=\)	\(\mathbf{3.5}\)

Equations with Brackets

📖 Theory

How to solve an equation with brackets

Worded problems

Common pitfalls

Frequently asked questions

Theory