Adding and subtracting algebraic fractions – Algebraic Fractions (Senior 3 Mathematics) | Ganzaa

Algebraic Fractions

Lessons

Senior 3 · Mathematics

Algebraic Fractions

5 lessons

Lesson 3 of 5

Adding and subtracting algebraic fractions

24 min

One fraction instead of two

A cyclist rides 12 km up a hill at an average speed of $v\ \mathrm{km\,h^{-1}}$ , then 12 km back down, 6 km per hour faster. The total time, in hours, is

\frac{12}{v} + \frac{12}{v + 6}.

To work with this total, for example to find the speed that gives a time of 2 hours, you need it as a single fraction. Adding algebraic fractions uses the same idea as $\frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}$ : rewrite both fractions over a common denominator, then add the numerators.

Definition

Lowest common denominator (LCD) of algebraic fractions: the simplest expression that every denominator divides exactly. It is the product of every factor that appears in the denominators, each raised to the highest power in which it appears.

Notation: factorise every denominator first, then build the LCD factor by factor.

Example: for the denominators $6 x^{2}$ and $4 x y$ , the LCD is $12 x^{2} y$ ; for $(x - 2) (x + 2)$ and $x + 2$ , it is $(x - 2) (x + 2)$ .

Important

Rules. With every denominator non-zero,

\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}, \qquad \frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}, \qquad \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}.

Why the third rule: multiplying a fraction by $\frac{d}{d} = 1$ does not change it, so $\frac{a}{b} = \frac{ad}{bd}$ and $\frac{c}{d} = \frac{bc}{bd}$ . Now both have the denominator $b d$ , and the first rule adds them.

The method

Factorise every denominator and write the conditions of existence.
Find the LCD.
Rewrite each fraction over the LCD by multiplying its numerator by the factors its denominator is missing.
Add or subtract the numerators, putting brackets round each numerator.
Simplify the numerator, factorise it if possible, and cancel any common factor.

Worked example 1: the same denominator

Simplify $\frac{3x}{x - 2} - \frac{6}{x - 2}$ .

Solution Condition: $x \ne 2$ .

\frac{3x}{x - 2} - \frac{6}{x - 2} = \frac{3x - 6}{x - 2} = \frac{3(x - 2)}{x - 2} = 3, \qquad x \ne 2.

Always look for a factor to cancel at the end.

Worked example 2: monomial denominators

Simplify $\frac{5}{6x^2} + \frac{3}{4xy}$ .

Solution Conditions: $x \ne 0$ , $y \ne 0$ . The LCD is $12 x^{2} y$ : the first denominator is missing $2 y$ , the second is missing $3 x$ .

\frac{5 \times 2y}{12x^2y} + \frac{3 \times 3x}{12x^2y} = \frac{10y + 9x}{12x^2y}.

Worked example 3: binomial denominators

Simplify $\frac{2}{x + 3} + \frac{5}{x - 1}$ .

Solution Conditions: $x \ne -3$ , $x \ne 1$ . The LCD is $(x + 3) (x - 1)$ .

\frac{2(x - 1) + 5(x + 3)}{(x + 3)(x - 1)} = \frac{2x - 2 + 5x + 15}{(x + 3)(x - 1)} = \frac{7x + 13}{(x + 3)(x - 1)}.

Check with $x = 0$ : $\frac{2}{3} + \frac{5}{-1} = -\frac{13}{3}$ , and $\frac{13}{(3)(-1)} = -\frac{13}{3}$ . ✓

Worked example 3 as a picture: each fraction is multiplied, top and bottom, by the factor of the LCD that its own denominator is missing.

Worked example 4: subtracting, with a factorised denominator

Simplify $\frac{x + 5}{x^2 - 4} - \frac{3}{x + 2}$ .

Solution $x^{2} - 4 = (x - 2) (x + 2)$ , so the conditions are $x \ne \pm 2$ and the LCD is $(x - 2) (x + 2)$ . The second fraction is missing the factor $x - 2$ :

\frac{(x + 5) - 3(x - 2)}{(x - 2)(x + 2)} = \frac{x + 5 - 3x + 6}{(x - 2)(x + 2)} = \frac{11 - 2x}{(x - 2)(x + 2)}.

The minus sign multiplies the whole of $3 (x - 2)$ , which gives $- 3 x + 6$ .

Worked example 5: the cyclist

Write the total time $\frac{12}{v} + \frac{12}{v + 6}$ as a single fraction, and use it to find the time when $v = 12$ .

Solution The LCD is $v (v + 6)$ , with $v \gt 0$ because $v$ is a speed.

\frac{12(v + 6) + 12v}{v(v + 6)} = \frac{24v + 72}{v(v + 6)} = \frac{24(v + 3)}{v(v + 6)}.

When $v = 12$ : $\frac{24 \times 15}{12 \times 18} = \frac{360}{216} = \frac{5}{3}$ hours, that is 1 hour 40 minutes. Check: $\frac{12}{12} + \frac{12}{18} = 1 + \frac{2}{3} = \frac{5}{3}$ . ✓

Worked example 6: denominators that are opposites

Simplify $\frac{4}{x - 3} + \frac{x}{3 - x}$ .

Solution Condition: $x \ne 3$ . Since $3 - x = - (x - 3)$ , $\frac{x}{3 - x} = -\frac{x}{x - 3}$ . So

\frac{4}{x - 3} - \frac{x}{x - 3} = \frac{4 - x}{x - 3}, \qquad x \ne 3.

Warning

Two costly errors. (1) Forgetting brackets when subtracting: $\frac{x + 5}{D} - \frac{3(x - 2)}{D}$ is $\frac{x + 5 - 3x + 6}{D}$ , not $\frac{x + 5 - 3x - 6}{D}$ . (2) Adding denominators: $\frac{1}{x} + \frac{1}{y}$ is not $\frac{2}{x + y}$ . Test with $x = y = 1$ : the left side is 2, the right side is 1. The correct sum is $\frac{x + y}{xy}$ .

Tip

Factorise before you multiply. Multiplying all the denominators together always gives a common denominator, but it may not be the lowest one, and the numerators then grow large. Factorising first gives the LCD and keeps the algebra short.

Khan Academy: Adding rational expression: unlike denominators (5 min). Watch how each fraction is multiplied by a form of 1 to reach the common denominator before the numerators are added.

✦

Practice

Write each as a single fraction in simplest form, with its conditions. The last three are harder.

$\frac{2x}{x + 4} + \frac{8}{x + 4}$
$\frac{3}{4a} + \frac{5}{6a^2}$
$\frac{1}{x} + \frac{1}{x + 2}$
$\frac{4}{x - 1} - \frac{3}{x + 2}$
$\frac{x}{x^2 - 9} + \frac{2}{x + 3}$
$\frac{5}{x - 2} + \frac{3}{2 - x}$
$\frac{2}{x^2 - x - 6} - \frac{1}{x^2 - 9}$
Show that $\frac{1}{n} - \frac{1}{n + 1} = \frac{1}{n(n + 1)}$ . Use it to find $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4}$ .

Show answersHide answers

$\frac{2x + 8}{x + 4} = \frac{2(x + 4)}{x + 4} = 2$ , $x \ne -4$
$\frac{9a + 10}{12a^2}$ , $a \ne 0$
$\frac{2x + 2}{x(x + 2)} = \frac{2(x + 1)}{x(x + 2)}$ , $x \ne 0,\ x \ne -2$
$\frac{4(x + 2) - 3(x - 1)}{(x - 1)(x + 2)} = \frac{x + 11}{(x - 1)(x + 2)}$ , $x \ne 1,\ x \ne -2$
$\frac{x + 2(x - 3)}{(x - 3)(x + 3)} = \frac{3(x - 2)}{(x - 3)(x + 3)}$ , $x \ne \pm 3$
$\frac{5}{x - 2} - \frac{3}{x - 2} = \frac{2}{x - 2}$ , $x \ne 2$
$\frac{2(x + 3) - (x + 2)}{(x - 3)(x + 2)(x + 3)} = \frac{x + 4}{(x - 3)(x + 2)(x + 3)}$ , $x \ne 3,\ -2,\ -3$
$\frac{(n + 1) - n}{n(n + 1)} = \frac{1}{n(n + 1)}$ . The sum is $\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) = 1 - \frac{1}{4} = \frac{3}{4}$ .

Sources and credits

8.3: Add and Subtract Rational Expressions with a Common Denominator, from Elementary Algebra by Lynn Marecek, MaryAnne Anthony-Smith and Andrea Honeycutt Mathis, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 30 September 2026). Adapted by Ganzaa.
8.4: Add and Subtract Rational Expressions with Unlike Denominators, from Elementary Algebra by Lynn Marecek, MaryAnne Anthony-Smith and Andrea Honeycutt Mathis, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 30 September 2026). Adapted by Ganzaa.
The cyclist problem and the worked examples. Written by Ganzaa.
Video: Adding rational expression: unlike denominators, Khan Academy, embedded from YouTube.

The diagram, the practice items and the quiz questions were written by Ganzaa.

1One fraction instead of two

2The method

One fraction instead of two

The method