This page provides detailed, step-by-step solutions to questions on adding, subtracting, and simplifying rational expressions. You will find clear explanations, basic examples, and advanced practice problems designed to help students and teachers master this algebra topic.
Tip: You can use our calculator to simplify rational expressions to check your results!
Adding, subtracting, and simplifying rational expressions is done in the exact same way as adding and subtracting numerical fractions. There are two main cases:
Case 1: Same Denominator
If the rational expressions already have the same denominator, simply add or subtract the numerators and keep the denominator the same.
$$ \frac{A}{C} \pm \frac{B}{C} = \frac{A \pm B}{C} $$
Case 2: Different Denominators
If the expressions have different denominators, you must first find the Lowest Common Multiple (LCM) to convert them to a common denominator before combining the numerators.
Subtract and simplify: $$ \dfrac{2}{3} - \dfrac{4}{3} $$
The two fractions have the same denominator, so we subtract the numerators directly:
$$ \dfrac{2}{3} - \dfrac{4}{3} = \dfrac{2-4}{3} = - \dfrac{2}{3} $$
Subtract and simplify: $$ \dfrac{1}{5} - \dfrac{7}{10} $$
The two fractions have different denominators, so we need to find the lowest common multiple (LCM) of 5 and 10.
We convert the fractions to the common denominator of 10:
$$ \dfrac{1}{5} - \dfrac{7}{10} = \dfrac{1 \times \color{red}{2}}{5 \times \color{red}{2}} - \dfrac{7}{10} = \dfrac{2}{10} - \dfrac{7}{10} $$
Now simplify:
$$ = \dfrac{2-7}{10} = - \dfrac{5}{10} = - \dfrac{1}{2} $$
Subtract and simplify: $$ \dfrac{2-x}{x+2} - \dfrac{4}{x+2} $$
The two rational expressions have the same denominator, so we subtract the numerators and keep the same denominator:
$$ \dfrac{2-x}{x+2} - \dfrac{4}{x+2} = \dfrac{2-x-4}{x+2} $$
Simplify the numerator:
$$ = \dfrac{-x-2}{x+2} $$
Factor out a $-1$ from the numerator and cancel the common factor:
$$ = \dfrac{-(x+2)}{x+2} = -1 \quad \text{for } x \neq -2 $$
For the following questions, you will need to know how to find the LCM of Expressions.
Evaluate: $$ \frac{7}{6} + \frac{1}{18} - \frac{5}{24} $$
Find the LCM of the denominators 6, 18, and 24.
Convert all 3 denominators to the common denominator 72:
$$ \frac{7 \times \color{red}{12}}{6 \times \color{red}{12}} + \frac{1 \times \color{red}{4}}{18 \times \color{red}{4}} - \frac{5 \times \color{red}{3}}{24 \times \color{red}{3}} $$
$$ = \frac{84}{72} + \frac{4}{72} - \frac{15}{72} = \frac{84 + 4 - 15}{72} = \frac{73}{72} $$
Simplify: $$ \frac{x+3}{x+5} + \frac{x-3}{x+2} $$
The two denominators $(x + 5)$ and $(x + 2)$ have no common factors, hence their LCM is given by:
$$ \text{LCM} = (x + 5)(x + 2) $$
Rewrite the rational expressions with the same denominator:
$$ \frac{(x+3)\color{red}{(x+2)}}{(x+5)\color{red}{(x+2)}} + \frac{(x-3)\color{red}{(x+5)}}{(x+2)\color{red}{(x+5)}} $$
$$ = \frac{(x+3)(x+2) + (x-3)(x+5)}{(x+5)(x+2)} $$
Expand, simplify, and factor the numerator if possible:
$$ = \frac{x^2 + 5x + 6 + x^2 + 2x - 15}{(x+5)(x+2)} = \frac{2x^2 + 7x - 9}{(x+5)(x+2)} $$
$$ = \frac{(x-1)(2x+9)}{(x+5)(x+2)} $$
Simplify: $$ \frac{-3}{x-4} + x + 4 $$
To add a rational expression to an expression without a denominator, multiply the whole expression by the common denominator over itself:
$$ \frac{-3}{x-4} + (x+4) \cdot \frac{\color{red}{x-4}}{\color{red}{x-4}} $$
The two expressions now have the same denominator and are added as follows:
$$ = \frac{-3 + (x+4)(x-4)}{x-4} = \frac{-3 + x^2 - 16}{x-4} = \frac{x^2 - 19}{x-4} $$
Simplify: $$ \frac{x-4}{x^2 - 3x + 2} - \frac{x+5}{x^2 + 2x - 3} $$
Factor the denominators completely to find the LCM:
$$ x^2 - 3x + 2 = (x - 1)(x - 2) $$
$$ x^2 + 2x - 3 = (x - 1)(x + 3) $$
$$ \text{LCM} = (x - 1)(x - 2)(x + 3) $$
Rewrite the rational expressions with the common denominator:
$$ \frac{x-4}{(x-1)(x-2)} \cdot \frac{\color{red}{x+3}}{\color{red}{x+3}} - \frac{x+5}{(x-1)(x+3)} \cdot \frac{\color{red}{x-2}}{\color{red}{x-2}} $$
$$ = \frac{(x-4)(x+3)}{(x-1)(x-2)(x+3)} - \frac{(x+5)(x-2)}{(x-1)(x+3)(x-2)} $$
Subtract the numerators, expand, and simplify:
$$ = \frac{(x-4)(x+3) - (x+5)(x-2)}{(x-1)(x-2)(x+3)} $$
$$ = \frac{(x^2 - x - 12) - (x^2 + 3x - 10)}{(x-1)(x-2)(x+3)} = \frac{-4x - 2}{(x-1)(x-2)(x+3)} $$
Simplify: $$ \frac{x^2 - x - 6}{x^2 - 2x - 3} - \frac{x^2 + 5x - 6}{x^2 + 9x + 18} $$
Rewrite the given expression with numerators and denominators in factored form and cancel common factors first:
$$ \frac{(x - 3)(x + 2)}{(x - 3)(x + 1)} - \frac{(x + 6)(x - 1)}{(x + 6)(x + 3)} = \frac{x + 2}{x + 1} - \frac{x - 1}{x + 3} $$
The two new denominators $(x + 1)$ and $(x + 3)$ have no common factors, so their LCM is $(x + 1)(x + 3)$. Rewrite with the common denominator:
$$ = \frac{(x + 2)(x + 3) - (x - 1)(x + 1)}{(x + 1)(x + 3)} $$
Expand and simplify:
$$ = \frac{(x^2 + 5x + 6) - (x^2 - 1)}{(x + 1)(x + 3)} = \frac{5x + 7}{(x + 1)(x + 3)} $$
Simplify: $$ \frac{2}{2x - 1} + \frac{x + 8}{2x^2 + 9x - 5} - \frac{5}{2x + 10} $$
Factor completely the three denominators to find the LCM:
$$ 2x - 1 = 2x - 1 $$
$$ 2x^2 + 9x - 5 = (2x - 1)(x + 5) $$
$$ 2x + 10 = 2(x + 5) $$
$$ \text{LCM} = 2(2x - 1)(x + 5) $$
Rewrite the rational expressions with the common denominator:
$$ \frac{2}{2x-1} \cdot \frac{\color{red}{2(x+5)}}{\color{red}{2(x+5)}} + \frac{x+8}{(2x-1)(x+5)} \cdot \frac{\color{red}{2}}{\color{red}{2}} - \frac{5}{2(x+5)} \cdot \frac{\color{red}{(2x-1)}}{\color{red}{(2x-1)}} $$
Combine the numerators and simplify:
$$ = \frac{2 \cdot 2(x+5) + 2(x+8) - 5(2x-1)}{2(2x-1)(x+5)} $$
$$ = \frac{4x + 20 + 2x + 16 - 10x + 5}{2(2x-1)(x+5)} $$
$$ = \frac{-4x + 41}{2(2x-1)(x+5)} = \frac{-(4x-41)}{2(2x-1)(x+5)} $$
Simplify: $$ \frac{y-2}{y(xy-y+3x-3)} + \frac{x}{2x-2} $$
Factor completely the two denominators to find the LCM:
$$ y(xy - y + 3x - 3) = y(y(x - 1) + 3(x - 1)) = y(x - 1)(y + 3) $$
$$ 2x - 2 = 2(x - 1) $$
$$ \text{LCM} = 2y(x - 1)(y + 3) $$
Rewrite the rational expressions with the common denominator:
$$ \frac{y-2}{y(x-1)(y+3)} \cdot \frac{\color{red}{2}}{\color{red}{2}} + \frac{x}{2(x-1)} \cdot \frac{\color{red}{y(y+3)}}{\color{red}{y(y+3)}} $$
$$ = \frac{2(y-2) + xy(y+3)}{2y(x-1)(y+3)} $$
Expand and simplify:
$$ = \frac{2y - 4 + xy^2 + 3xy}{2y(x-1)(y+3)} = \frac{xy^2 + 3xy + 2y - 4}{2y(x-1)(y+3)} $$