Equations with Algebraic Fractions
Equations with algebraic fractions are easiest after multiplying every term by the lowest common denominator to clear the fractions. This NSW Year 9 Advanced maths subtopic starts with one fraction (NESA outcome MA5‑EQU‑C‑01) and moves on to equations with two or more fractions (MA5‑EQU‑P‑02).
You will solve equations such as \(\frac{3 - 2x}{12} + 3 = 7\), \(\frac{2d - 5}{9} = \frac{d - 3}{6}\) and \(\frac{7 + 2m}{5} - \frac{1 - m}{2} = -1\), taking care with the sign in front of a fraction. You will write equations for fractions of a number and of Priya's money, find an unknown \(k\) from a given solution, and show that one equation has no solution. It is part of Year 9 Advanced Equations.
Theory
To solve an equation with algebraic fractions, multiply every term on both sides by the lowest common multiple of the denominators. The fractions disappear and an ordinary linear equation is left. This Year 9 Advanced maths page covers one, two and three fractions, a subtracted fraction, fraction word problems and checking the answer.
An algebraic fraction has a pronumeral in the numerator, such as \(\dfrac{x + 3}{4}\). On this page every denominator is a number.
The quickest way to deal with fractions in an equation is to remove them all at once. Multiply every term on both sides by the lowest common multiple (LCM) of the denominators, the smallest number they all divide into. Each fraction then cancels to a whole expression.
A fraction bar groups its numerator, so put brackets round each numerator as the fraction clears: \(12 \times \dfrac{x - 2}{6} = 2(x - 2)\). This matters most when a fraction is subtracted, because the minus sign then applies to every term of its numerator.
Terms without a fraction are multiplied too. After clearing, expand, collect like terms and solve as usual. If the \(x\) terms cancel and leave a false statement, the equation has no solution.
Clearing two fractions with LCM \(L\):
| Denominators | LCM | Each fraction becomes |
|---|---|---|
| \(2\) and \(5\) | \(10\) | \(5(\ldots)\) and \(2(\ldots)\) |
| \(4\) and \(10\) | \(20\) | \(5(\ldots)\) and \(2(\ldots)\) |
| \(3\), \(4\) and \(6\) | \(12\) | \(4(\ldots)\), \(3(\ldots)\) and \(2(\ldots)\) |
How to solve an equation with algebraic fractions
- Find the LCM of all the denominators.
- Multiply every term on both sides by it. Cancel each fraction and put brackets round its numerator. Multiply terms without fractions too.
- Expand the brackets carefully, especially after a minus sign.
- Collect like terms and solve the linear equation.
- Check by substituting into the original equation, working out each side separately.
Shortcuts and special cases
- One fraction plus a number: \(\dfrac{x + 1}{4} + 3 = 12\) can also be solved by subtracting \(3\) first, then multiplying by \(4\).
- Same denominators: if every fraction has denominator \(3\), multiply every term by \(3\).
- Word problems: "one-fifth of a number" is \(\dfrac{x}{5}\). "Two-thirds of what is left" needs the amount left written first, then multiplied by \(\dfrac{2}{3}\).
- A constant to find: substitute the given solution for \(x\) first, then clear fractions and solve for the constant.
- No solution: if the \(x\) terms cancel and leave something false like \(4 = 7\), say so.
Multiply both sides by \(5\). The whole right side is multiplied, so it needs brackets.
| \(2x - 1\) | \(=\) | \(5(x - 4)\) |
| \(2x - 1\) | \(=\) | \(5x - 20\) |
| \(19\) | \(=\) | \(3x\) |
| \(x\) | \(=\) | \(\mathbf{\dfrac{19}{3}}\) |
Check: \(2x - 1 = \dfrac{35}{3}\), so LHS \(= \dfrac{7}{3}\); RHS \(= \dfrac{19}{3} - 4 = \dfrac{7}{3}\).
The lowest common multiple of \(4\) and \(6\) is \(12\). Multiply every term by \(12\), with brackets round each numerator.
| \(3(x + 3) - 2(x - 2)\) | \(=\) | \(24\) |
| \(3x + 9 - 2x + 4\) | \(=\) | \(24\) |
| \(x + 13\) | \(=\) | \(24\) |
| \(x\) | \(=\) | \(\mathbf{11}\) |
The minus in front of the second fraction makes \(-2 \times (-2) = +4\). Check: \(\dfrac{14}{4} - \dfrac{9}{6} = 3.5 - 1.5 = 2\).
The lowest common multiple of \(3\), \(4\) and \(6\) is \(12\).
| \(4x + 3(x - 2)\) | \(=\) | \(2(x + 8)\) |
| \(7x - 6\) | \(=\) | \(2x + 16\) |
| \(5x\) | \(=\) | \(22\) |
| \(x\) | \(=\) | \(\mathbf{4.4}\) |
Check: LHS \(= \dfrac{4.4}{3} + \dfrac{2.4}{4}\) \(= 1.4\overline{6} + 0.6\) \(= 2.0\overline{6}\); RHS \(= \dfrac{12.4}{6} = 2.0\overline{6}\).
Let his pay be \(\$x\). Rent is \(\dfrac{x}{3}\) and food is \(\dfrac{x}{4}\).
| \(x - \dfrac{x}{3} - \dfrac{x}{4}\) | \(=\) | \(350\) |
| \(12x - 4x - 3x\) | \(=\) | \(4200 \quad (\times 12)\) |
| \(5x\) | \(=\) | \(4200\) |
| \(x\) | \(=\) | \(840\) |
His weekly pay is \(\mathbf{\$840}\). Check: rent \(\$280\), food \(\$210\), and \(840 - 280 - 210 = 350\).
Common pitfalls
Frequently asked questions
How do you solve equations with fractions?
Multiply every term on both sides by the lowest common multiple of the denominators. This removes the fractions. Then expand any brackets, collect like terms and solve the linear equation.
What is the lowest common multiple of the denominators?
It is the smallest number that every denominator divides into exactly. For denominators 4 and 6 it is 12; for 3 and 5 it is 15.
Why do you need brackets when clearing fractions?
The fraction bar groups the whole numerator. When the fraction is multiplied out, every term of the numerator must be multiplied, and if the fraction is subtracted every term changes sign. Brackets keep that straight.
What is cross-multiplying?
When one fraction equals another, \(\dfrac{a}{p} = \dfrac{b}{q}\), multiplying both sides by \(pq\) gives \(qa = pb\). It is a shortcut for clearing both fractions at once.
Can an equation with fractions have no solution?
Yes. If after clearing the fractions the \(x\) terms cancel and leave a false statement such as \(5 = 1\), no value of \(x\) works.