Variables on Both Sides
When the pronumeral appears on both sides, first move all the pronumeral terms to one side and the numbers to the other. This NSW Year 9 Advanced maths subtopic solves these linear equations, including ones with decimals and fractions (NESA outcome MA5‑EQU‑C‑01).
You will solve equations such as \(10 - 6a = -25 - 11a\) and \(\frac{2x + 1}{3} = \frac{x + 4}{2}\), giving fraction answers where needed. You will use the equal sides of a rectangle, compare two hire companies, find when two tanks hold the same amount, find an unknown \(k\), and explain why \(2x + 7 = 2x - 3\) has no solution while another equation is true for every \(x\). It is part of Year 9 Advanced Equations.
Theory
When the pronumeral is on both sides of an equation, move all the pronumeral terms to one side and the numbers to the other, then solve. This Year 9 Advanced maths page covers negative and decimal coefficients, fractions on both sides, break-even problems where two costs are equal, and equations with no solution or every value as a solution.
In an equation such as \(5x + 2 = 2x + 11\) the pronumeral appears on both sides. Adding or subtracting the same pronumeral term on both sides keeps the equation balanced, just as with numbers.
Remove the smaller pronumeral term from both sides (or add the more negative one). The pronumeral term left over is then positive, which avoids dividing by a negative.
When there are fractions, multiply every term on both sides by the lowest common multiple of the denominators. Each numerator with more than one term goes in brackets.
Most linear equations have exactly one solution. If the pronumeral terms cancel and leave a false statement, such as \(9 = -1\), there is no solution. If they leave a true statement, such as \(6 = 6\), every value is a solution and the equation is an identity.
On a graph, the solution of \(ax + b = cx + d\) is the \(x\)-value where the lines \(y = ax + b\) and \(y = cx + d\) meet. Parallel lines never meet (no solution); the same line meets everywhere (identity).
Collecting terms, for \(a \ne c\):
| After collecting the \(x\) terms | Number of solutions |
|---|---|
| \(kx = m\) with \(k \ne 0\) | one: \(x = \dfrac{m}{k}\) |
| a false statement, e.g. \(5 = 2\) | none |
| a true statement, e.g. \(5 = 5\) | infinitely many (identity) |
How to solve with the pronumeral on both sides
- Clear fractions first, if any: multiply every term by the lowest common multiple of the denominators.
- Move the pronumeral terms to one side by subtracting the smaller one (or adding the negative one) on both sides.
- Move the numbers to the other side, then divide by the coefficient.
- Check by working out each side separately with your answer.
Problems that lead to both sides
- Two plans or two tanks: write each quantity in terms of the same pronumeral (fixed part \(+\) rate \(\times\) time, or start \(-\) rate \(\times\) time) and set them equal. Answer with the time and, if asked, the common value.
- Geometry: equal lengths (opposite sides of a rectangle, equal sides of an isosceles triangle) give two equal expressions.
- A given solution: if the solution is \(x = 2k - 1\), substitute it for \(x\); you get an equation in \(k\) to solve the same way.
- No solution or identity: if the pronumeral disappears, decide whether what is left is true or false.
The \(x\) terms are \(-4x\) and \(2x\). Add \(4x\) to both sides so the \(x\) term left over is positive.
| \(9\) | \(=\) | \(6x - 15\) |
| \(24\) | \(=\) | \(6x\) |
| \(x\) | \(=\) | \(\mathbf{4}\) |
Check: LHS \(= 9 - 16 = -7\), RHS \(= 8 - 15 = -7\).
Multiply every term by \(15\), the lowest common multiple of \(3\) and \(5\). Keep each numerator in brackets.
| \(5(x + 2)\) | \(=\) | \(3(2x - 1) + 15\) |
| \(5x + 10\) | \(=\) | \(6x - 3 + 15\) |
| \(5x + 10\) | \(=\) | \(6x + 12\) |
| \(-2\) | \(=\) | \(x\) |
So \(x = \mathbf{-2}\). Check: LHS \(= \dfrac{0}{3} = 0\), RHS \(= \dfrac{-5}{5} + 1 = 0\).
Let \(w\) be the number of weeks. Write each total cost and set them equal.
| \(60 + 14w\) | \(=\) | \(20 + 18w\) |
| \(40\) | \(=\) | \(4w\) |
| \(w\) | \(=\) | \(\mathbf{10}\) |
After \(10\) weeks both cost \(60 + 14(10) = 20 + 18(10) = \$200\). After that, Gym B costs \(\$4\) a week more, so Gym A is cheaper.
(i) \(7x - 3 = 2x + 12\)
(ii) \(4x + 9 = 4x - 1\)
(iii) \(0.5x + 3 = \dfrac{x + 6}{2}\)
Collect the \(x\) terms on one side and see what is left.
(i)
| \(5x\) | \(=\) | \(15\) |
| \(x\) | \(=\) | \(3\) |
One solution, \(x = 3\).
(ii) Subtract \(4x\) from both sides:
| \(9\) | \(=\) | \(-1\) |
The \(x\) terms cancel and the statement is false for every \(x\): no solution.
(iii) Multiply both sides by \(2\):
| \(x + 6\) | \(=\) | \(x + 6\) |
Both sides are the same expression, so it is true for every value of \(x\) (an identity).
Common pitfalls
Frequently asked questions
How do you solve equations with variables on both sides?
Subtract the smaller variable term from both sides so the variable is only on one side. Then move the numbers to the other side and divide by the coefficient. Check by substituting into both sides.
Which side should the x terms go on?
Either side works. Moving them to the side with the larger x term keeps the coefficient positive, which avoids dividing by a negative number.
What does it mean if an equation has no solution?
It means no value of the pronumeral makes both sides equal. The \(x\) terms cancel and leave a false statement such as \(3 = 8\). On a graph the two sides are parallel lines that never meet.
What is an identity?
An identity is an equation that is true for every value of the pronumeral, because both sides simplify to the same expression, such as \(2(x + 3) = 2x + 6\).
How do you find when two phone plans cost the same?
Write each plan's cost as fixed charge \(+\) rate \(\times\) amount used, set the two expressions equal and solve. Substitute back to find the common cost.