Solving one-sided linear equations
Use inverse operations to make the unknown stand alone. Apply each operation to both sides. Multiplication and division preserve equivalent equations only when the multiplier or divisor is non-zero
\[\begin{gathered}ax+b=c,\quad a\ne0\\x=\frac{c-b}{a}\end{gathered}\]
- Undo addition or subtraction In this example, subtract 7 from both sides, then divide both sides by negative 2. Keep the sign of the coefficient \(\displaystyle \begin{aligned}7-2x&=20\\-2x&=13\\x&=-\frac{13}{2}\end{aligned}\)
- A constant outside a fraction Add 6 to both sides, then multiply both sides by 5. Multiplying first is also valid if every term on each side is multiplied \(\displaystyle \begin{aligned}\frac{x}{5}-6&=3\\\frac{x}{5}&=9\\x&=45\end{aligned}\)
- A constant inside the numerator Multiply both sides by 5, subtract 9 from both sides, then divide both sides by 2. The fraction bar covers the whole numerator \(\displaystyle \begin{aligned}\frac{2x+9}{5}&=7\\2x+9&=35\\2x&=26\\x&=13\end{aligned}\)
- A scaled bracket Divide both sides by 7, then add 2 to both sides. Dividing by the non-zero outside factor avoids expansion here \(\displaystyle \begin{aligned}7(x-2)&=42\\x-2&=6\\x&=8\end{aligned}\)
- Check the original equation Substitute the solution into the original equation. Equal values on the two sides confirm the solution
Worked example
Solve the equation and check the solution
\[\frac{5x}{4}+3=-4\]
- \(\displaystyle \frac{5x}{4}=-7\)Subtract 3 from both sides
- \(\displaystyle 5x=-28\)Multiply both sides by 4
- \(\displaystyle x=-\frac{28}{5}\)Divide both sides by 5
- \(\displaystyle \text{Left side}=\frac{5(-\frac{28}{5})}{4}+3\)Substitute negative twenty-eight fifths into the original left side
- \(\displaystyle \text{Left side}=\frac{-28}{4}+3\)Calculate the numerator
- \(\displaystyle \text{Left side}=-7+3\)Divide before adding the outside constant
- \(\displaystyle \text{Left side}=-4\)Complete the addition
- \(\displaystyle \text{Right side}=-4\)Both original sides have the same value, so the solution is verified
Common mistakes
Adding instead of undoing addition
✕\(\displaystyle \begin{aligned}5x+6&=41\\5x&=47\end{aligned}\)✓\(\displaystyle \begin{aligned}5x+6&=41\\5x&=35\end{aligned}\)
Subtract the added constant from both sidesLosing the negative coefficient
✕\(\displaystyle \begin{aligned}-6x&=21\\x&=\frac{7}{2}\end{aligned}\)✓\(\displaystyle \begin{aligned}-6x&=21\\x&=-\frac{7}{2}\end{aligned}\)
Divide by the signed coefficient. Dividing a positive number by a negative number gives a negative resultMultiplying only part of a side
✕\(\displaystyle \begin{aligned}\frac{x}{6}+2&=9\\x+2&=54\end{aligned}\)✓\(\displaystyle \begin{aligned}\frac{x}{6}+2&=9\\x+12&=54\end{aligned}\)
When multiplying both sides by the denominator, multiply the outside constant too. Subtracting the outside constant first is another valid routeDividing again to clear a denominator
✕\(\displaystyle \begin{aligned}\frac{2x-5}{4}&=7\\2x-5&=\frac{7}{4}\end{aligned}\)✓\(\displaystyle \begin{aligned}\frac{2x-5}{4}&=7\\2x-5&=28\end{aligned}\)
Undo division by multiplying both sides by the denominatorDifficulty levels
introBegin with one inverse operation, then solve two-step equations, including signed whole-number solutions
standardDivide first for simple scaled brackets. Solve both whole-numerator and outside-constant fraction forms, including exact fractional solutions
examChoose an exact solution using the same one-sided equation forms
Questions about this topic
Must I always subtract before dividing?+
No. Undoing the outside operation first often gives a short route. If you choose a different valid operation, apply it to every term on both sides
Can a solution be negative or fractional?+
Yes. A value is a solution when substitution makes both original sides equal
Must I expand a bracket?+
No. When a factor multiplies the whole bracket, you can divide both sides by that non-zero factor first
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 substitutionTranslate between a description and an algebraic expressionSimplify algebraic terms and collect like termsSimplify 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)