This page provides detailed, step-by-step solutions to questions on adding, subtracting, and simplifying rational expressions . You will find clear explanations, worked examples, and practice problems designed to help students, teachers, and parents master this important algebra topic.
\[ \frac{7}{6} + \frac{1}{18} - \frac{5}{24} \]
The three denominators in the fractions above are different and therefore we need to find a common denominator.
We first find the lowest common multiple (LCM) of the two denominators 6, 18 and 24.
6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 80,...
18: 18, 36, 54, 72, 90,...
24: 24, 48, 72, 96...
The lowest common denominator is 72 and we now convert all 3 denominators to the common denominator 72
\[ \frac{7}{6} + \frac{1}{18} - \frac{5}{24} = \frac{7 \times \textcolor{red}{12}}{6 \times \textcolor{red}{12}} + \frac{1 \times \textcolor{red}{4}}{18 \times \textcolor{red}{4}} - \frac{5 \times \textcolor{red}{3}}{24 \times \textcolor{red}{3}} \]
and simplify as follows:
\[ = \frac{84}{72} + \frac{4}{72} - \frac{15}{72} = \frac{84 + 4 - 15}{72} = \frac{73}{72} \]
\[ \frac{x+3}{x+5} + \frac{x-3}{x+2} \]
The two rational expressions have different denominators. In order to add the rational expressions above, we need to convert them to a common denominator. The two denominators \( x + 5 \) and \( x + 2 \) have no common factors hence their LCM is given by:
\[ \text{LCM} = (x + 5)(x + 2) \]
We now use the LCM as the common denominator and rewrite the rational expressions with the same denominator as follows.
\[ \frac{x+3}{x+5} + \frac{x-3}{x+2} = \frac{(x+3)\textcolor{red}{(x+2)}}{(x+5)\textcolor{red}{(x+2)}} + \frac{(x-3)\textcolor{red}{(x+5)}}{(x+2)\textcolor{red}{(x+5)}} \] \[ = \frac{(x+3)(x+2) + (x-3)(x+5)}{(x+5)(x+2)} \]
We now expand, simplify and factor the numerator if possible.
\[ = \frac{x^2 + 5x + 6 + x^2 + 2x - 15}{(x+5)(x+2)} = \frac{2x^2 + 7x - 9}{(x+5)(x+2)} \] \[ = \frac{(x-1)(2x+9)}{(x+5)(x+2)} \]
\[ \frac{-3}{x-4} + x + 4 \]
In order to add a rational expression with an expression without denominator, we convert the one without denominator into a rational expression then add them.
\[ \frac{-3}{x-4} + x + 4 = \frac{-3}{x-4} + (x+4)\cdot\frac{\textcolor{red}{x-4}}{\textcolor{red}{x-4}} \]
The two rational expressions have the same denominator and they are added as follows:
\[ = \frac{-3 + (x+4)(x-4)}{x-4} = \frac{x^2 - 19}{x-4} \]
\[ \frac{x-4}{x^2 - 3x + 2} - \frac{x+5}{x^2 + 2x - 3} \]
The two rational expressions have different denominators. In order to add the rational expressions above, we need to convert them to a common denominator. We first factor completely the two denominators \( x^2 - 3x + 2 \) and \( x^2 + 2 x - 3 \) and find the LCM of Expressions.
\[ x^2 - 3x + 2 = (x - 1) (x - 2) \] \[ x^2 + 2 x - 3 = (x - 1)(x + 3) \] \[ \text{LCM} = (x - 1)(x - 2)(x + 3) \]
We now use the LCM as the common denominator and rewrite the rational expressions with the same denominator as follows.
\[ \frac{x-4}{x^2-3x+2} - \frac{x+5}{x^2+2x-3} = \frac{x-4}{(x-1)(x-2)} - \frac{x+5}{(x-1)(x+3)} \] \[ = \frac{x-4}{(x-1)(x-2)} \cdot \frac{x+3}{x+3} - \frac{x+5}{(x-1)(x+3)} \cdot \frac{x-2}{x-2} \] \[ = \frac{(x-4)(x+3)}{(x-1)(x-2)(x+3)} - \frac{(x+5)(x-2)}{(x-1)(x+3)(x-2)} \]
We now add the numerators expand and simplify.
\[ = \frac{(x-4)(x+3) - (x+5)(x-2)}{(x-1)(x-2)(x+3)} = \frac{x^2 - x - 12 - (x^2 + 3x - 10)}{(x-1)(x-2)(x+3)} = \frac{-4x - 2}{(x-1)(x-2)(x+3)} \]
\[ \frac{x^2 - x - 6}{x^2 - 2x - 3} - \frac{x^2 + 5x - 6}{x^2 + 9x + 18} \]
We rewrite the given expression with numerators and denominators in factored form and simplify if possible.
\[ \frac{x^2 - x - 6}{x^2 - 2x - 3} - \frac{x^2 + 5x - 6}{x^2 + 9x + 18} = \frac{(x - 3)(x + 2 )}{(x - 3)(x +1)} - \frac{(x + 6)(x - 1)}{(x + 6)(x + 3)} \]
We cancel common factors.
\[ = \frac{\cancel{(x - 3)}(x + 2)}{\cancel{(x - 3)}(x + 1)} - \frac{\cancel{(x + 6)}(x - 1)}{\cancel{(x + 6)}(x + 3)} = \frac{x + 2}{x + 1} - \frac{x - 1}{x + 3} \]
The two denominators \( x + 1 \) and \( x + 3 \) have no common factors and therefore their LCD is \( (x + 1)(x + 3) \). We rewrite the above with the common factor \( (x + 1)(x + 3) \) as follows:
\[ = \frac{(x + 2)(x + 3) - (x - 1)(x + 1)}{(x + 1)(x + 3)} \] Expand and simplify. \[ = \frac{5x + 7}{(x + 1)(x + 3)} \]
\[ \frac{2}{2x - 1} + \frac{x + 8}{2x^2 + 9x - 5} - \frac{5}{2x + 10} \]
The three rational expressions have different denominators. In order to subtract/add the rational expressions above, we need to convert them to a common denominator. List and factor completely the three denominators \( 2x - 1 \), \( 2 x^2 + 9 x - 5 \) and \( 2 x + 10 \) and find the LCM.
\[ 2x - 1 = 2x - 1 \]
\[ 2 x^2 + 9 x - 5 = (2x - 1)(x + 5) \]
\[ 2x+10 = 2(x + 5) \]
\[ \text{LCM} = 2(2x - 1)(x + 5) \]
We now use the LCM as the common denominator and rewrite the rational expressions with the same denominator as follows.
\[ \frac{2}{2x-1} + \frac{x+8}{2x^2+9x-5} - \frac{5}{2x+10} \] \[ = \frac{2}{2x-1} \cdot \frac{\color{red}2(x+5)}{\color{red}2(x+5)} + \frac{x+8}{(2x-1)(x+5)} \cdot \frac{\color{red}2}{\color{red}2} - \frac{5}{2(x+5)} \cdot \frac{\color{red}(2x-1)}{\color{red}(2x-1)} \]
We now add the numerators and simplify.
\[ = \frac{2 \cdot 2(x+5) + 2(x+8) - 5(2x-1)}{2(2x-1)(x+5)} = \frac{-(4x-41)}{2(2x-1)(x+5)} \]
\[ \frac{y-2}{y(xy-y+3x-3)} + \frac{x}{2x-2} \]
The two rational expressions have different denominators. In order to subtract/add the rational expressions above, we need to convert them to a common denominator. List and factor completely the two denominators \( y(x y - y + 3 x - 3) \) and \( 2 x - 2 \) and find the LCM.
\[ y(x y - y + 3 x - 3) = y( y(x - 1) + 3 (x - 1)) = y(x - 1)(y + 3) \]
\[ 2 x - 2 = 2(x - 1) \]
\[ \text{LCM} = 2 y (x - 1)(y + 3) \]
We now use the LCM as the common denominator and rewrite the rational expressions with the same denominator as follows.
\[ \frac{y-2}{y(xy-y+3x-3)}+\frac{x}{2x-2}=\frac{y-2}{y(x-1)(y+3)}+\frac{x}{2(x-1)} \] \[ =\frac{y-2}{y(x-1)(y+3)}\cdot\frac{\color{red}2}{\color{red}2}+\frac{x}{2(x-1)}\cdot\frac{\color{red}y(y+3)}{\color{red}y(y+3)} \] \[ =\frac{2(y-2)+xy(y+3)}{2y(x-1)(y+3)} \]
Expand and simplify.
\[ =\frac{xy^{2}+3xy+2y-4}{2y(x-1)(y+3)} \]