Resources For Teachers For Tutors For Students & Parents AI New Pricing
Year 9 Maths Linear and Simultaneous Equations

Equations with Brackets and Pronumerals on Both Sides

20 practice questions 1 video lesson Theory + worked examples
Practice 20 questions
Practice questions

Every question with a fully worked solution.

Start practising
Watch 1 video(s)
  • Equations with Brackets and Pronumerals on Both Sides - Video - Solve a Linear Equation with Parentheses and Variables on Both Sides: -a(x+b)+c=dx+e Watch
Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

To solve an equation with brackets, expand every bracket first, collect the like terms, then solve as usual. This Year 9 maths page shows how to expand correctly, including a negative or a bare minus in front of a bracket, two brackets, pronumerals on both sides, and a bracket over a denominator.

To expand a bracket, multiply every term inside by the term outside. \(3(x + 4)\) becomes \(3x + 12\).

A negative outside changes the sign of every term inside. \(-2(x - 5)\) becomes \(-2x + 10\).

A bare minus in front of a bracket means \(-1\). \(-(x + 3)\) becomes \(-x - 3\).

Like terms have the same pronumeral part. Collect them on each side, such as \(4n + 3n = 7n\).

Then solve the equation exactly as before: gather the pronumerals on one side, the numbers on the other, and divide.

Area model for expanding a bracketA rectangle 3 high is split into two parts, x wide and 4 wide. The parts have areas 3x and 12, so 3 times the bracket x plus 4 equals 3x plus 12. 3x 12 x 4 3 3(x + 4) = 3x + 12
The number outside multiplies every term inside: \(3 \times x\) and \(3 \times 4\).
Expanding a bracket with a negative in frontThe expression negative 3 times the bracket x minus 4. Red arrows show negative 3 multiplying x and multiplying negative 4. The result is negative 3x plus 12. βˆ’3 ( x βˆ’ 4 ) (βˆ’3) Γ— x = βˆ’3x (βˆ’3) Γ— (βˆ’4) = +12 = βˆ’3x + 12
A negative outside changes the sign of every term inside. \((-3) \times (-4) = +12\).

Expanding a bracket:

\[a(b + c) = ab + ac \qquad a(b - c) = ab - ac\]
a(b+c)=ab+ac
BracketExpands toNote
\(5(x + 2)\)\(5x + 10\)both terms multiplied
\(5(x - 2)\)\(5x - 10\)the sign travels with the \(2\)
\(-5(x - 2)\)\(-5x + 10\)both signs change
\(-(x - 2)\)\(-x + 2\)a bare minus is \(-1\)
Pronumerals on both sides. Take the smaller pronumeral term off both sides, so the one that is left is positive. In \(7 - 4x = 1 - 6x\), \(-6x\) is smaller, so add \(6x\) to both sides: \(7 + 2x = 1\), so \(2x = -6\) and \(x = -3\).
A bracket over a denominator. Multiply both sides by the denominator first, then expand. \(\dfrac{2(x + 1)}{3} = 4\) becomes \(2(x + 1) = 12\), so \(2x + 2 = 12\) and \(x = 5\).
Fractions inside a bracket. Expand as usual and simplify each product. \(6\left(\dfrac{x}{3} - \dfrac{1}{2}\right) = 2x - 3\), because \(6 \times \dfrac{x}{3} = 2x\) and \(6 \times \dfrac{1}{2} = 3\).
Check by substituting into the original equation, brackets and all.

How to solve an equation with brackets

  1. Clear any denominator under a bracket by multiplying both sides by it.
  2. Expand every bracket. Watch the signs when the number outside is negative.
  3. Collect like terms on each side.
  4. Solve: take the smaller pronumeral term off both sides, move the numbers to the other side, then divide. Check by substituting.
Example 1 β€” One bracket
Solve
(i) \(3(x + 4) = 27\)
(ii) \(-(x - 6) = 3x - 2\).
Solution

(i) Expand the bracket, then solve as usual.

\(3x + 12\)\(=\)\(27\)
\(3x\)\(=\)\(15\)
\(x\)\(=\)\(\boldsymbol{5}\)
x=5

(ii) A bare minus means \(-1\). It changes both signs inside.

\(-x + 6\)\(=\)\(3x - 2\)
\(6\)\(=\)\(4x - 2\)
\(8\)\(=\)\(4x\)
\(x\)\(=\)\(\boldsymbol{2}\)
x=2

Check (ii): LHS \(= -(2 - 6) = 4\). RHS \(= 6 - 2 = 4\). \(\checkmark\)

Example 2 β€” Pronumerals on both sides
Solve \(4(2y - 3) = 3y + 13\).
Solution

Expand, then take the smaller pronumeral term, \(3y\), off both sides.

\(8y - 12\)\(=\)\(3y + 13\)
\(5y - 12\)\(=\)\(13\)
\(5y\)\(=\)\(25\)
\(y\)\(=\)\(\boldsymbol{5}\)
y=5

Check: LHS \(= 4(10 - 3) = 28\). RHS \(= 15 + 13 = 28\). \(\checkmark\)

Example 3 β€” Two brackets
Solve \(4(n - 1) + 3(n + 2) = 30\).
Solution

Expand both brackets. Collect the like terms before moving anything.

\(4n - 4 + 3n + 6\)\(=\)\(30\)
\(7n + 2\)\(=\)\(30\)
\(7n\)\(=\)\(28\)
\(n\)\(=\)\(\boldsymbol{4}\)
n=4

Check: \(4 \times 3 + 3 \times 6 = 12 + 18 = 30\) \(\checkmark\)

Example 4 β€” Subtracting a bracket
Solve \(2(3w + 1) - 5(w - 4) = 30\).
Solution

The second bracket is multiplied by \(-5\). So \(-5 \times w = -5w\) and \(-5 \times (-4) = +20\).

\(6w + 2 - 5w + 20\)\(=\)\(30\)
\(w + 22\)\(=\)\(30\)
\(w\)\(=\)\(\boldsymbol{8}\)
w=8

Check: \(2 \times 25 - 5 \times 4 = 50 - 20 = 30\) \(\checkmark\)

Writing \(-20\) instead of \(+20\) is the most common mistake here.

Common pitfalls

Multiply the second term too. \(3(x + 4)\) is \(3x + 12\), not \(3x + 4\).
Take a minus all the way through. \(-2(x - 3)\) is \(-2x + 6\). The second sign changes as well as the first.
Watch for a bracket being subtracted. In \(10 - (3 - x)\), the minus belongs to the whole bracket: \(10 - 3 + x\).
Collect before you move. With two brackets, expand both and collect like terms on each side before taking anything across.

Frequently asked questions

How do you solve equations with brackets?

Expand every bracket, collect like terms on each side, then solve: pronumerals to one side, numbers to the other, then divide.

What happens when there is a negative in front of a bracket?

The negative multiplies every term inside, so every sign inside changes. \(-3(x - 4) = -3x + 12\).

What does a minus sign in front of a bracket mean?

A bare minus means \(-1\) times the bracket. It changes the sign of every term inside: \(-(x + 3) = -x - 3\).

Do you have to expand the brackets first?

Usually, yes. If the bracket sits over a denominator, multiply by the denominator first. If the number outside divides the other side exactly, you can divide first instead: \(4(x - 2) = 20\) gives \(x - 2 = 5\).

How do you check an equation with brackets?

Substitute your answer into the original equation, brackets and all. Work out each side. They should be equal.