Variables on Both Sides
When the pronumeral appears on both sides, the new first step is to gather those terms on one side. After that the equation is solved just like a one-sided one. This NSW Year 9 Core maths subtopic is Stage 5 Core content under NESA outcome MA5‑EQU‑C‑01.
You will move the smaller pronumeral term to avoid working with negatives, as in \(5x + 2 = 3x + 10\), handle equations where the pronumeral term ends up negative, such as \(15 - 7w = 22 - 3w\), and give answers as fractions or decimals when they do not come out whole. You will also clear fractions on one or both sides, as in \(\frac{2x + 1}{3} = \frac{x + 4}{2}\), and check by substituting into the original equation. Equations like these compare two phone plans or two savings plans. It is part of Year 9 Core Equations.
Theory
When the pronumeral is on both sides of an equation, gather the pronumeral terms onto one side first. After that it is an ordinary equation. This Year 9 maths page shows which term to move, how to handle negative terms, and how to clear fractions on both sides.
In \(5x + 2 = x + 14\) the pronumeral \(x\) appears on both sides. Gather all the pronumeral terms on one side and all the numbers on the other.
Take a term off by doing the opposite to both sides. Take \(x\) off by subtracting \(x\). Take \(-3x\) off by adding \(3x\).
Move the smaller pronumeral term. Then the term left over has a positive coefficient, and there is no dividing by a negative at the end.
Answers can be negative, decimals or fractions. Write a fraction in its simplest form.
Check by substituting the answer into the original equation. Work out each side on its own. They must match.
The first move removes the smaller pronumeral term:
| Equation | First move | Leaves |
|---|---|---|
| \(6x = 2x + 12\) | subtract \(2x\) from both sides | \(4x = 12\) |
| \(3x + 5 = x + 11\) | subtract \(x\), then subtract \(5\) | \(2x = 6\) |
| \(10 - 3x = 2x\) | add \(3x\) to both sides | \(10 = 5x\) |
A fraction on each side: multiply every term by a number both denominators divide into.
One fraction equals one fraction: cross-multiply.
How to solve an equation with pronumerals on both sides
- Clear any fractions first, by multiplying every term by a common denominator (or cross-multiplying).
- Collect the pronumeral terms: take the smaller one off both sides.
- Move the numbers to the other side, then divide by the coefficient.
- Check by substituting into the original equation.
Take the smaller pronumeral term, \(3x\), off both sides. Then move the number.
| \(7x - 4 - 3x\) | \(=\) | \(3x + 12 - 3x\) |
| \(4x - 4\) | \(=\) | \(12\) |
| \(4x\) | \(=\) | \(16\) |
| \(x\) | \(=\) | \(\boldsymbol{4}\) |
Check: LHS \(= 7 \times 4 - 4 = 24\). RHS \(= 3 \times 4 + 12 = 24\). \(\checkmark\)
The term \(-2k\) is subtracted, so add \(2k\) to both sides. That leaves a positive \(6k\).
| \(9 - 2k + 2k\) | \(=\) | \(4k - 3 + 2k\) |
| \(9\) | \(=\) | \(6k - 3\) |
| \(12\) | \(=\) | \(6k\) |
| \(k\) | \(=\) | \(\boldsymbol{2}\) |
Check: LHS \(= 9 - 4 = 5\). RHS \(= 8 - 3 = 5\). \(\checkmark\)
(i) \(5p + 8 = 2p - 10\)
(ii) \(13 - 8n = 3 - 2n\).
(i) Take \(2p\) off both sides, then take \(8\) off both sides.
| \(3p + 8\) | \(=\) | \(-10\) |
| \(3p\) | \(=\) | \(-18\) |
| \(p\) | \(=\) | \(\boldsymbol{-6}\) |
(ii) Both terms are subtracted. Add \(8n\) to both sides, which removes the smaller term, \(-8n\), so a positive term is left.
| \(13\) | \(=\) | \(3 + 6n\) |
| \(10\) | \(=\) | \(6n\) |
| \(n\) | \(=\) | \(\dfrac{10}{6}\) |
| \(n\) | \(=\) | \(\boldsymbol{\dfrac{5}{3}}\) |
Simplify the fraction. The answer can be negative or a fraction.
(i) \(\dfrac{x}{3} + 2 = \dfrac{x}{6} + 5\)
(ii) \(\dfrac{x + 3}{4} = \dfrac{2x - 1}{3}\).
(i) Both \(3\) and \(6\) divide into \(6\). Multiply every term by \(6\).
| \(2x + 12\) | \(=\) | \(x + 30\) |
| \(x + 12\) | \(=\) | \(30\) |
| \(x\) | \(=\) | \(\boldsymbol{18}\) |
(ii) One fraction equals one fraction, so cross-multiply. Put brackets round each top.
| \(3(x + 3)\) | \(=\) | \(4(2x - 1)\) |
| \(3x + 9\) | \(=\) | \(8x - 4\) |
| \(13\) | \(=\) | \(5x\) |
| \(x\) | \(=\) | \(\boldsymbol{\dfrac{13}{5}}\) |
As a decimal, \(x = 2.6\).
Common pitfalls
Frequently asked questions
How do you solve equations with variables on both sides?
Take the smaller pronumeral term off both sides so all the pronumerals are on one side. Then move the numbers to the other side and divide by the coefficient.
Which side should the variable go on?
Either side works. Moving the smaller pronumeral term leaves a positive coefficient, which avoids dividing by a negative.
How do you get rid of a negative x term?
Add it to both sides. For \(9 - 2k\), add \(2k\) to both sides and the term disappears from that side.
How do you solve an equation with fractions on both sides?
Multiply every term by a number that all the denominators divide into. If it is one fraction equal to one fraction, cross-multiply.
Can the answer to an equation be a fraction?
Yes. If \(6n = 10\), then \(n = \dfrac{10}{6} = \dfrac{5}{3}\). Leave it as a simplified fraction, or write it as a decimal if asked.