Equations with brackets and unknowns on both sides
Use expansion from 2C and inverse operations from 2D. Collect like terms on each side before removing an unknown term from both sides. Finish with inverse operations, dividing only by a non-zero number
\[a(bx+c)=abx+ac\]
- Include the sign of the outside factor Multiply 2x and −3 by −6. The two negative factors give a positive constant of 18 \(\displaystyle -6(2x-3)=-12x+18\)
- Collect terms on their own side first Multiply 2x and 5 by 3, then combine 6x and −4x on the left. Subtract 15 from both sides and divide by 2 \(\displaystyle \begin{aligned}3(2x+5)-4x&=23\\6x+15-4x&=23\\2x+15&=23\\2x&=8\\x&=4\end{aligned}\)
- Apply the same subtraction to both sides Subtract 4x from both sides to leave 5x on the left. Add 7, then divide both sides by 5 \(\displaystyle \begin{aligned}9x-7&=4x+18\\5x-7&=18\\5x&=25\\x&=5\end{aligned}\)
- Choose a useful first step The whole left side is multiplied by 6, so divide both sides by 6, then subtract 4. Expansion is also valid. If a term sits outside a bracket, retain it \(\displaystyle \begin{aligned}6(x+4)&=42\\x+4&=7\\x&=3\end{aligned}\)
Worked example
Solve the equation and check the solution
\[5(2x-1)=3(x+3)\]
- \(\displaystyle 10x-5=3(x+3)\)Multiply 2x and −1 by 5. Keep the right side unchanged
- \(\displaystyle 10x-5=3x+9\)Multiply x and 3 by 3
- \(\displaystyle 7x-5=9\)Subtract 3x from both sides
- \(\displaystyle 7x=14\)Add 5 to both sides
- \(\displaystyle x=2\)Divide both sides by 7
- \(\displaystyle \text{Left}=5(2\times2-1)\)Substitute 2 into the original left side
- \(\displaystyle \text{Left}=5(4-1)\)Multiply 2 by 2 inside the bracket first
- \(\displaystyle \text{Left}=5\times3\)Subtract 1 from 4 inside the bracket
- \(\displaystyle \text{Left}=15\)Multiply 5 by 3
- \(\displaystyle \text{Right}=3(2+3)\)Substitute 2 into the original right side
- \(\displaystyle \text{Right}=3\times5\)Add 2 and 3 inside the bracket
- \(\displaystyle \text{Right}=15\)Both original sides have the same value, so the solution is correct
Common mistakes
Losing the sign of a product
✕\(\displaystyle -5(2x-4)=-10x-20\)✓\(\displaystyle -5(2x-4)=-10x+20\)
Multiplying −5 by −4 gives positive 20. Multiply both terms in the bracket by −5Changing only one side
✕\(\displaystyle \begin{aligned}8x+3&=5x-6\\3x+3&=5x-6\end{aligned}\)✓\(\displaystyle \begin{aligned}8x+3&=5x-6\\3x+3&=-6\end{aligned}\)
Subtract 5x from both sides. This removes 5x from the right and leaves 3x on the leftDropping an outside term
✕\(\displaystyle \begin{aligned}4(x+5)-3x&=29\\4x+20&=29\end{aligned}\)✓\(\displaystyle \begin{aligned}4(x+5)-3x&=29\\4x+20-3x&=29\end{aligned}\)
Expanding the bracket gives 4x and 20. Keep the outside term −3x, then combine it with 4xDifficulty levels
standardSolve equations with brackets, including signed factors and like terms to collect
challengeSolve equations with unknowns on both sides, then combine this with brackets. This is core section work
examChoose exact solutions to bracket, bracket-and-collection and both-sides equations from four options
Questions about this topic
Must I always expand first?+
No. If a whole side is one scaled bracket, dividing by its non-zero outside factor can be shorter. Expand when it helps you collect terms. Both methods must preserve the whole equation
Which unknown term should I subtract?+
Either can work. Removing the smaller positive coefficient often leaves a positive coefficient. Removing the other term is still valid, but may leave a negative coefficient
Where should I check my answer?+
Substitute into both sides of the original equation. Checking only a later line can miss an earlier expansion or sign error
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 9 Mathematics, chapter 2: Linear equations
In this chapter
Translate between a description and an algebraic expressionEvaluate an algebraic expression by substitutionSimplify algebraic terms and collect like termsExpand brackets and collect like terms2D Two-step linear equations: Solve a linear equation2E Linear equations with brackets: Solve a linear equationSet up and solve a practical linear equationSolve and interpret a linear inequationRepresent and read inequalities on a number lineSubstitute values into a formulaSolve for an unknown or recognise a rearranged formulaSolve a pair of simultaneous linear equations (optional extension)Solve a pair of simultaneous linear equations by elimination (optional extension)Set up a pair of simultaneous equations from a description (optional extension)Solve a practical problem with simultaneous equations (optional extension)