Equations with Algebraic Fractions
Theory
To solve an equation with fractions, multiply every term by the lowest common denominator so the fractions disappear, then solve the equation that is left. This Year 10A maths page (Victorian Curriculum 10A) covers numerical and pronumeral denominators, binomial denominators, excluded values and solutions that must be rejected, changing the subject, and rate problems. Curriculum: VC2M10A12 / VC2M10AA03 / AC9M10A01.
An equation stays balanced if you do the same thing to both sides. Multiplying every term by the lowest common denominator (LCD) clears all the fractions at once. For \(\dfrac{x}{4}+\dfrac{x}{6} = 5\) the LCD is \(12\), giving \(3x+2x = 60\).
When there is exactly one fraction on each side, cross-multiplying is a shortcut for the same step: \(\dfrac{a}{b} = \dfrac{c}{d}\) becomes \(ad = bc\).
If the pronumeral is in a denominator, some values are excluded: they make a denominator zero, so the fraction does not exist. In \(\dfrac{2}{x-5} = 1\), \(x \ne 5\). An answer equal to an excluded value is rejected, and the equation may have no solution.
With binomial denominators the \(x^2\) terms often cancel after expanding, leaving a linear equation. Always check the answer by substituting it into the original equation.
Which LCD to multiply by:
| Denominators | Multiply every term by |
|---|---|
| \(4,\ 6,\ 3\) | \(12\) |
| \(x,\ 5,\ 2x\) | \(10x\) |
| \(3x,\ 4x,\ 6\) | \(12x\) |
| \(x+1,\ x-2\) | \((x+1)(x-2)\) |
| \(x-4,\ x+4,\ x^2-16\) | \((x-4)(x+4)\) |
How to solve an equation with fractions
- Exclude: write down any value that makes a denominator zero. Factorise quadratic denominators to find them.
- Simplify first if it helps: fractions with the same denominator on opposite sides can be collected onto one side.
- Clear the fractions: multiply every term on both sides by the LCD, or cross-multiply if there is one fraction on each side.
- Expand the brackets. A minus sign in front of a fraction multiplies its whole numerator.
- Solve the linear equation. If \(x^2\) terms appear, they should cancel.
- Check the answer is not excluded, then substitute it into the original equation.
Word problems
- Time, distance, speed: time \(= \dfrac{\text{distance}}{\text{speed}}\). Write a fraction for each part of the trip, then use the fact the question gives (total time, or a difference in times). Convert minutes to hours first: \(20\) min \(= \dfrac{1}{3}\) h.
- Work rates: add the rates of everything helping, subtract the rate of anything working against, and set the total equal to the combined rate (Example 4).
- Answer the question asked. If you solve for a speed but the question wants the time, substitute back.
The LCD of \(4\), \(3\) and \(6\) is \(12\). Multiply every term on both sides by \(12\).
| \(12\times\dfrac{3x-2}{4}-12\times\dfrac{x+1}{3}\) | \(=\) | \(12\times\dfrac{x}{6}\) |
| \(3(3x-2)-4(x+1)\) | \(=\) | \(2x\) |
| \(9x-6-4x-4\) | \(=\) | \(2x\) |
| \(5x-10\) | \(=\) | \(2x\) |
| \(3x\) | \(=\) | \(10\) |
| \(x\) | \(=\) | \(\boldsymbol{\dfrac{10}{3}}\) |
The bracket in \(-4(x+1)\) makes both terms negative: \(-4x-4\).
(i) \(\dfrac{3}{x}+\dfrac{1}{4}=\dfrac{5}{2x}\)
(ii) \(\dfrac{x+5}{x-2}=\dfrac{x+1}{x-4}\)
(i) \(x \ne 0\). The LCD of \(x\), \(4\) and \(2x\) is \(4x\).
| \(4x\times\dfrac{3}{x}+4x\times\dfrac{1}{4}\) | \(=\) | \(4x\times\dfrac{5}{2x}\) |
| \(12+x\) | \(=\) | \(10\) |
| \(x\) | \(=\) | \(\boldsymbol{-2}\) |
Check: \(-\dfrac{3}{2}+\dfrac{1}{4} = -\dfrac{5}{4}\) and \(\dfrac{5}{-4} = -\dfrac{5}{4}\).
(ii) \(x \ne 2\) and \(x \ne 4\). One fraction on each side, so cross-multiply.
| \((x+5)(x-4)\) | \(=\) | \((x+1)(x-2)\) |
| \(x^2+x-20\) | \(=\) | \(x^2-x-2\) |
| \(2x\) | \(=\) | \(18\) |
| \(x\) | \(=\) | \(\boldsymbol{9}\) |
The \(x^2\) terms cancel, leaving a linear equation. Check: \(\dfrac{14}{7} = 2\) and \(\dfrac{10}{5} = 2\).
Factorise \(x^2-9 = (x-3)(x+3)\). So \(x \ne 3\) and \(x \ne -3\). Multiply every term by \((x-3)(x+3)\).
| \(3(x+3)-(x-3)\) | \(=\) | \(18\) |
| \(3x+9-x+3\) | \(=\) | \(18\) |
| \(2x+12\) | \(=\) | \(18\) |
| \(x\) | \(=\) | \(3\) |
But \(x = 3\) makes \(x-3\) zero, so it is not allowed. The equation has no solution.
Multiplying by an expression that can be zero can create a false solution. Always test your answer against the excluded values.
Let pipe B take \(x\) hours. In one hour A fills \(\dfrac{1}{10}\) of the pool, B fills \(\dfrac{1}{x}\), and together they fill \(\dfrac{1}{6}\).
| \(\dfrac{1}{10}+\dfrac{1}{x}\) | \(=\) | \(\dfrac{1}{6}\) |
| \(\dfrac{1}{x}\) | \(=\) | \(\dfrac{1}{6}-\dfrac{1}{10}\) |
| \(\) | \(=\) | \(\dfrac{5}{30}-\dfrac{3}{30} = \dfrac{1}{15}\) |
| \(x\) | \(=\) | \(\mathbf{15}\text{ hours}\) |
Check: \(\dfrac{1}{10}+\dfrac{1}{15} = \dfrac{3}{30}+\dfrac{2}{30} = \dfrac{1}{6}\). Pipe B is slower than A, which makes sense.
Common pitfalls
Frequently asked questions
How do you solve equations with fractions?
Find the lowest common denominator of every fraction in the equation, multiply every term on both sides by it, then solve the equation that is left. Check the answer in the original equation.
When can you cross-multiply?
Only when the equation is one single fraction equal to another single fraction, \(\dfrac{a}{b} = \dfrac{c}{d}\). Then \(ad = bc\). With more terms, multiply by the lowest common denominator instead.
What do you do when x is in the denominator?
Note the values of \(x\) that make a denominator zero, because they are not allowed. Then multiply every term by the lowest common denominator, which includes \(x\) or the brackets, and solve as usual.
Can an equation with fractions have no solution?
Yes. If the only value you find makes a denominator zero, it must be rejected, so the equation has no solution. This is why you always check your answer against the excluded values.
How do you solve work rate problems?
If something does a job in \(t\) hours, its rate is \(\dfrac{1}{t}\) of the job per hour. Add the rates of everything working together, subtract any that work against the job, and set the total equal to \(\dfrac{1}{\text{combined time}}\).