Equations with Fractions
Here the numbers either side of the equals sign can be fractions too, as in \(\frac{x}{5} - \frac{1}{2} = \frac{1}{2}\). The method in NSW Year 8 maths (NESA outcome MA4‑EQU‑C‑01) stays the same: undo one operation per line on both sides, using a common denominator whenever fractions are added or subtracted.
You will solve equations with a grouped top line such as \(\frac{x + 3}{5} = -\frac{1}{2}\), mixed numerals, and fraction terms on both sides that combine, and see that \(\frac{4x}{5} = \frac{2}{3}\) can be started by multiplying or by dividing. Answers are simplified and checked by substitution. You will also use a formula with a fraction line, such as finding a missing number from a mean. It is part of Year 8 Equations.
Every question with a fully worked solution.
- Equations with Fractions - Video - Equations with Fractions Watch
Theory
Fractions change nothing about the method: undo one operation per line, on both sides. When there are several fractions, multiply every term by the lowest common denominator (LCD) to clear them: \(\dfrac{3x}{5} + \dfrac{x}{15} = 7\) becomes \(9x + x = 105\), so \(x = 10.5\). This Year 8 maths page shows how.
A fraction can be added to or taken from both sides, exactly like a whole number. If the denominators differ, rewrite both fractions over a common denominator first.
A mixed numeral is often easier as an improper fraction: \(2\dfrac{1}{4} = \dfrac{9}{4}\). In \(p + 2\dfrac{1}{4} = 5\), subtract \(2\dfrac{1}{4}\) from both sides: \(p = 2\dfrac{3}{4}\).
A fraction line groups. In \(\dfrac{x + 1}{4} = -\dfrac{3}{2}\), multiply both sides by \(4\) first: \(x + 1 = -6\), so \(x = -7\).
Two steps, in either order. \(\dfrac{ax}{b} = c\) can be undone by multiplying by \(b\) first or by dividing by \(a\) first. Both give the same solution.
Clear all the fractions with the LCD. Find the lowest common denominator of every fraction, then multiply each term on both sides by it. Each denominator cancels, leaving an equation with whole numbers.
Fraction terms on both sides. Like terms combine: \(\dfrac{5a}{4} - \dfrac{a}{4} = \dfrac{4a}{4} = a\). With different denominators, clear them with the LCD first: \(\dfrac{x}{2} + 1 = \dfrac{x}{5} + 4\) times \(10\) gives \(5x + 10 = 2x + 40\), so \(x = 10\).
Simplify the answer, then write it as a mixed numeral or a terminating decimal: \(\dfrac{9}{5} = 1\dfrac{4}{5} = 1.8\).
Clearing fractions with the LCD:
| Equation | Multiply every term by the LCD |
|---|---|
| \(\dfrac{x}{2} + 1 = \dfrac{x}{5} + 4\) | \(\times 10\): \(5x + 10 = 2x + 40\), so \(x = 10\) |
| \(\dfrac{2x}{3} + \dfrac{1}{4} = 5\) | \(\times 12\): \(8x + 3 = 60\), so \(x = 7\dfrac{1}{8}\) |
| \(\dfrac{x - 2}{4} = x - 5\) | \(\times 4\): \(x - 2 = 4x - 20\), so \(x = 6\) |
One fraction
- Undo any loose number (adding or subtracting fractions over a common denominator).
- Multiply both sides by the denominator, then divide by the number in front.
- Simplify and check.
Several fractions
- Find the LCD of every fraction.
- Multiply every term on both sides by the LCD and cancel (put a grouped numerator in brackets).
- Solve the whole-number equation and check.
| \(\dfrac{y}{4}\) | \(=\) | \(\dfrac{4}{5} - \dfrac{2}{5} = \dfrac{2}{5}\) |
| \(y\) | \(=\) | \(\dfrac{2}{5} \times 4 = \dfrac{8}{5}\) |
| \(\) | \(=\) | \(1\dfrac{3}{5}\) |
\(y = 1\dfrac{3}{5}\). Check: \(\dfrac{8}{5} \div 4 + \dfrac{2}{5} = \dfrac{2}{5} + \dfrac{2}{5} = \dfrac{4}{5}\).
Add \(\dfrac{1}{2}\) to both sides, using a common denominator of \(6\).
| \(\dfrac{n}{6}\) | \(=\) | \(\dfrac{2}{6} + \dfrac{3}{6} = \dfrac{5}{6}\) |
| \(n\) | \(=\) | \(5\) |
\(n = 5\). Check: \(\dfrac{5}{6} - \dfrac{3}{6} = \dfrac{2}{6} = \dfrac{1}{3}\).
The LCD of \(3\) and \(4\) is \(12\). Multiply every term by \(12\).
| \(12 \times \dfrac{2x}{3} + 12 \times \dfrac{1}{4}\) | \(=\) | \(12 \times 5\) |
| \(8x + 3\) | \(=\) | \(60\) |
| \(8x\) | \(=\) | \(57\) |
| \(x\) | \(=\) | \(\dfrac{57}{8} = 7\dfrac{1}{8}\) |
\(x = 7\dfrac{1}{8}\)
\(5\) divides \(15\), so the LCD is \(15\). Multiply every term by \(15\).
| \(15 \times \dfrac{3x}{5} + 15 \times \dfrac{x}{15}\) | \(=\) | \(15 \times 7\) |
| \(9x + x\) | \(=\) | \(105\) |
| \(10x\) | \(=\) | \(105\) |
| \(x\) | \(=\) | \(10.5\) |
\(x = 10.5\)
(i) \(\dfrac{5a}{4} - \dfrac{1}{3} = \dfrac{a}{4} + \dfrac{2}{3}\)
(ii) \(\dfrac{x - 2}{4} = x - 5\)
(i) Subtract \(\dfrac{a}{4}\) from both sides: \(\dfrac{4a}{4} = a\).
(ii) Multiply both sides by \(4\); the right side becomes \(4(x - 5)\).
| \(a - \dfrac{1}{3}\) | \(=\) | \(\dfrac{2}{3}\) |
| \(a\) | \(=\) | \(1\) |
| \(x - 2\) | \(=\) | \(4x - 20\) |
| \(18\) | \(=\) | \(3x\) |
| \(x\) | \(=\) | \(6\) |
(i) \(a = 1\). Check: both sides equal \(\dfrac{11}{12}\).
(ii) \(x = 6\). Check: \(\dfrac{6 - 2}{4} = 1\) and \(6 - 5 = 1\).
Multiply by \(3\) first, then divide by \(2\); or divide by \(2\) first, then multiply by \(3\).
| \(2x\) | \(=\) | \(\dfrac{4}{9} \times 3 = \dfrac{4}{3}\) |
| \(x\) | \(=\) | \(\dfrac{4}{3} \div 2 = \dfrac{2}{3}\) |
| \(\dfrac{x}{3}\) | \(=\) | \(\dfrac{4}{9} \div 2 = \dfrac{2}{9}\) |
| \(x\) | \(=\) | \(\dfrac{2}{9} \times 3 = \dfrac{2}{3}\) |
Both orders give \(x = \dfrac{2}{3}\).
Common pitfalls
Frequently asked questions
How do you solve an equation with fractions?
Multiply every term on both sides by the lowest common denominator to clear the fractions, then solve as usual.
What is the lowest common denominator?
The smallest number that every denominator divides into: for \(3\) and \(4\) it is \(12\).
Can the answer be a fraction?
Yes. Simplify it and write it as a mixed numeral or decimal, such as \(\dfrac{9}{5} = 1.8\).
What do I do with a mixed numeral?
Change it to an improper fraction, such as \(1\dfrac{1}{2} = \dfrac{3}{2}\), or work with whole and fraction parts.
Why put brackets around a numerator?
The fraction line groups the whole top, so multiplying by the denominator leaves the whole top: \(4 \times \dfrac{x + 1}{4} = x + 1\).