How to Add Rational Expressions
How to Add Rational Expressions Example
Subtract Rational Expressions Example
Practice Problems
Problem 1
Step 1
Identify the common denominator.
In this case, the denominator is $$(x+5)(x-2)$$.
Step 2
Rewrite each expression so that it has the common denominator.
$$ \begin{align*} \frac 9 {x + 5} - \frac{11}{x - 2} & = \blue{\frac{x-2}{x-2}}\cdot \frac 9 {x + 5} - \frac{11}{x - 2}\cdot \red{\frac{x+5}{x+5}}\\[6pt] & = \frac{9\blue{(x-2)}}{(x + 5)\blue{(x-2)}} - \frac{11\red{(x+5)}}{\red{(x+5)}(x - 2)} \end{align*} $$
Step 3
Distribute in the numerator.
$$ \begin{align*} \frac{\blue{9(x-2)}}{(x + 5)(x-2)} - \frac{\red{11(x+5)}}{(x+5)(x - 2)} & = \frac{\blue{9x - 18}}{(x + 5)(x-2)} - \frac{\red{11x + 55}}{(x+5)(x - 2)} \end{align*} $$
Step 4
Subtract the numerators, but keep the denominator the same. Make sure you subtract the entire second numerator! (Use parentheses).
$$ \begin{align*} \frac{\blue{9x - 18}}{(x + 5)(x-2)} - \frac{\red{11x + 55}}{(x+5)(x - 2)} & = \frac{\blue{9x - 18} - \red{(11x + 55)}}{(x+5)(x - 2)}\\[5pt] & = \frac{\blue{9x - 18}\,\red{\,-11x - 55}}{(x+5)(x - 2)}\\[5pt] & = \frac{\blue{9x}\,\red{\,-11x}\,\blue{\,-18}\,\red{\,-55}}{(x+5)(x - 2)}\\[5pt] & = \frac{-2x-73}{(x+5)(x - 2)} \end{align*} $$
Step 5
(Optional) Some instructors expect their students to expand the denominator.
$$ \begin{align*} \frac{-2x-73}{(x+5)(x - 2)} & = \frac{-2x-73}{x^2 + 3x - 10} \end{align*} $$
Answer
$$ \begin{align*} \frac 9 {x + 5} - \frac{11}{x - 2} = \frac{-2x-73}{(x+5)(x - 2)} \end{align*} $$
Problem 2
Step 1
Identify the common denominator.
In this case, the common denominator is $$(x-6)(x+3)$$.
Step 2
Rewrite each expression so that it has the common denominator.
$$ \begin{align*} \frac 3 {x - 6} + \frac 7 {x + 3} & = \blue{\frac{x+3}{x+3}}\cdot\frac 3 {x - 6} + \frac 7 {x + 3}\cdot\red{\frac{x-6}{x-6}}\\[6pt] & = \frac{3\blue{(x+3)}}{(x - 6)\blue{(x+3)}} + \frac{7\red{(x-6)}}{\red{(x-6)}(x + 3)} \end{align*} $$
Step 3
Distribute the numerators.
$$ \begin{align*} \frac{3\blue{(x+3)}}{(x - 6)\blue{(x+3)}} + \frac{7\red{(x-6)}}{\red{(x-6)}(x + 3)} & = \frac{\blue{3x + 9}}{(x - 6)(x+3)} + \frac{\red{7x-42}}{(x-6)(x + 3)} \end{align*} $$
Step 4
Add the numerators, but keep the denominator the same.
$$ \begin{align*} \frac{\blue{3x + 9}}{(x - 6)(x+3)} + \frac{\red{7x-42}}{(x-6)(x + 3)} & = \frac{\blue{3x + 9} + \red{7x-42}}{(x-6)(x + 3)}\\[6pt] & = \frac{\blue{3x}\,+\,\red{7x} + \blue{9} - \red{42}}{(x-6)(x + 3)}\\[6pt] & = \frac{10x - 33}{(x-6)(x + 3)} \end{align*} $$
Step 5
(Optional) Some instructors will expect you to expand the denominator.
$$ \begin{align*} \frac{10x - 33}{x^2 - 3x -18} \end{align*} $$
Answer
$$ \begin{align*} \frac 3 {x - 6} + \frac 7 {x + 3} = \frac{10x - 33}{(x-6)(x + 3)} \end{align*} $$
Problem 3
Step 1
Identify the common denominator.
In this case, the common denominator is $$(2x+1)(2x-3)$$.
Step 2
Rewrite each expression so it has the common denominator.
$$ \begin{align*} \frac 8 {2x + 1} + \frac 4 {2x - 3} & = \blue{\frac{2x - 3}{2x - 3}}\cdot \frac 8 {2x + 1} + \frac 4 {2x - 3}\cdot \red{\frac{2x + 1}{2x+1}}\\[6pt] & = \frac{8\blue{(2x - 3)}}{(2x + 1)\blue{(2x - 3)}} + \frac{4\red{(2x+1)}} {\red{(2x+1)}(2x - 3)} \end{align*} $$
Step 3
Distribute the numerators.
$$ \begin{align*} \frac{8\blue{(2x - 3)}}{(2x + 1)\blue{(2x - 3)}} + \frac{4\red{(2x+1)}} {\red{(2x+1)}(2x - 3)} & = \frac{\blue{16x - 24}}{(2x + 1)(2x - 3)} + \frac{\red{8x+4}} {(2x+1)(2x - 3)} \end{align*} $$
Step 4
Add the numerators, but keep the same denominator.
$$ \begin{align*} \frac{\blue{16x - 24}}{(2x + 1)(2x - 3)} + \frac{\red{8x+4}} {(2x+1)(2x - 3)} & = \frac{\blue{16x - 24} + \red{8x+4}} {(2x+1)(2x - 3)}\\[6pt] & = \frac{\blue{16x} + \red{8x} - \blue{24} + \red{4}} {(2x+1)(2x - 3)}\\[6pt] & = \frac{24x - 20} {(2x+1)(2x - 3)} \end{align*} $$
Step 5
(Optional) Some instructors may require you to expand the denominator.
$$ \begin{align*} \frac{24x - 20} {(2x+1)(2x - 3)} = \frac{24x - 20} {4x^2 -4x - 3} \end{align*} $$
Answer
$$ \begin{align*} \frac 8 {2x + 1} + \frac 4 {2x - 3} = \frac{24x - 20} {(2x+1)(2x - 3)} \end{align*}$$
Problem 4
Step 1
Identify the common denominator.
In this case, the common denominator is $$(3x - 2)(4x + 1)$$.
Step 2
Rewrite each expression so it has the common denominator.
$$ \begin{align*} \frac{15}{3x - 2} - \frac 1 {4x + 1} & = \blue{\frac{4x+1}{4x+1}}\cdot\frac{15}{3x - 2} - \frac 1 {4x + 1}\cdot\red{\frac{3x-2}{3x-2}}\\[6pt] & = \frac{15\blue{(4x+1)}}{(3x - 2)\blue{(4x+1)}} - \frac{\red{3x-2}}{\red{(3x-2)}(4x + 1)} \end{align*} $$
Step 3
Expand the numerator.
$$ \begin{align*} \frac{15\blue{(4x+1)}}{(3x - 2)\blue{(4x+1)}} - \frac{\red{3x-2}}{\red{(3x-2)}(4x + 1)} & = \frac{\blue{60x+15}}{(3x - 2)(4x+1)} - \frac{\red{3x-2}}{(3x-2)(4x + 1)} \end{align*} $$
Step 4
Subtract the numerators, but keep the same denominator. Make sure you subtract the entire second numerator!
$$ \begin{align*} \frac{\blue{60x+15}}{(3x - 2)(4x+1)} - \frac{\red{3x-2}}{(3x-2)(4x + 1)} & = \frac{\blue{60x+15} - \red{(3x-2)}}{(3x-2)(4x + 1)}\\[6pt] & = \frac{\blue{60x+15} - \red{3x+ 2}}{(3x-2)(4x + 1)}\\[6pt] & = \frac{\blue{60x} - \red{3x} + \blue{15} + \red{2}}{(3x-2)(4x + 1)}\\[6pt] & = \frac{57x + 17}{(3x-2)(4x + 1)} \end{align*} $$
Step 5
(Optional) Some instructors may require you to expand the denominator.
$$ \begin{align*} \frac{57x + 17}{(3x-2)(4x + 1)} = \frac{57x + 17}{12x^2 - 5x - 2} \end{align*} $$
Answer
$$ \begin{align*} \frac{15}{3x - 2} - \frac 1 {4x + 1} = \frac{57x + 17}{(3x-2)(4x + 1)} \end{align*} $$
Problem 5
Step 1
Identify the common denominator.
In this case, the common denominator is $$(x-1)(x+2)$$.
Step 2
Rewrite the expressions so they have the common denominator.
$$ \begin{align*} \frac{x + 4 }{x - 1} + \frac{x - 4}{x + 2} & = \blue{\frac{x+2}{x+2}}\cdot\frac{x + 4 }{x - 1} + \frac{x - 4}{x + 2}\cdot\red{\frac{x-2}{x-1}}\\[6pt] & = \frac{\blue{(x+2)}(x + 4)}{\blue{(x+2)}(x - 1)} + \frac{(x - 4)\red{(x-1)}}{(x + 2)\red{(x-1)}} \end{align*} $$
Step 3
Expand the numerators.
$$ \begin{align*} \frac{\blue{(x+2)}(x + 4)}{\blue{(x+2)}(x - 1)} + \frac{(x - 4)\red{(x-1)}}{(x + 2)\red{(x-1)}} & = \frac{\blue{x^2 + 6x + 8}}{(x+2)(x - 1)} + \frac{\red{x^2 - 5x + 4}}{(x + 2)(x-1)} \end{align*} $$
Step 4
Add the numerators, but keep the same denominator.
$$ \begin{align*} \frac{\blue{x^2 + 6x + 8}}{(x+2)(x - 1)} + \frac{\red{x^2 - 5x + 4}}{(x + 2)(x-1)} & = \frac{\blue{x^2 + 6x + 8} + \red{x^2 - 5x + 4}}{(x + 2)(x-1)}\\[6pt] & = \frac{\blue{x^2} + \red{x^2} + \blue{6x} - \red{5x} + \blue{8} + \red{4}}{(x + 2)(x-1)}\\[6pt] & = \frac{2x^2 + x + 12}{(x + 2)(x-1)} \end{align*} $$
Step 5
(Optional) Some instructors require students to expand the denominator.
$$ \begin{align*} \frac{2x^2 + x + 12}{(x + 2)(x-1)} = \frac{2x^2 + x + 12}{x^2 + x - 2} \end{align*} $$
Answer
$$ \begin{align*} \frac{x + 4 }{x - 1} + \frac{x - 4}{x + 2} = \frac{2x^2 + x + 12}{(x + 2)(x-1)} \end{align*} $$
Problem 6
Step 1
Identify the common denominator.
In this case, the common denominator is $$(x + 5)(x + 6)$$.
Step 2
Rewrite the expressions so they have the common denominator.
$$ \begin{align*} \frac{x - 10 }{x + 5} - \frac{x - 3}{x + 6} & = \blue{\frac{x+6}{x+6}}\cdot\frac{x - 10 }{x + 5} - \frac{x - 3}{x + 6}\cdot\red{\frac{x+5}{x+5}}\\[6pt] & = \frac{\blue{(x+6)}(x - 10) }{\blue{(x+6)}(x + 5)} - \frac{(x - 3)\red{(x+5)}}{(x + 6)\red{(x+5)}} \end{align*} $$
Step 3
Expand the numerators.
$$ \begin{align*} \frac{\blue{(x+6)}(x - 10) }{\blue{(x+6)}(x + 5)} - \frac{(x - 3)\red{(x+5)}}{(x + 6)\red{(x+5)}} & = \frac{\blue{x^2 - 4x - 60}}{(x+6)(x + 5)} - \frac{\red{x^2 + 2x - 15}}{(x + 6)(x+5)} \end{align*} $$
Step 4
Subtract the numerators, but keep the same denominator. Be sure to subtract the entire second numerator!
$$ \begin{align*} \frac{\blue{x^2 - 4x - 60}}{(x+6)(x + 5)} - \frac{\red{x^2 + 2x - 15}}{(x + 6)(x+5)} & = \frac{\blue{x^2 - 4x - 60} - (\red{x^2 + 2x - 15})}{(x + 6)(x+5)}\\[6pt] & = \frac{\blue{x^2 - 4x - 60} - \red{x^2 - 2x + 15}}{(x + 6)(x+5)}\\[6pt] & = \frac{\blue{x^2} - \red{x^2} - \blue{4x} - \red{2x} - \blue{60} + \red{15}}{(x + 6)(x+5)}\\[6pt] & = \frac{- 6x - 45}{(x + 6)(x+5)} \end{align*} $$
Step 5
(Optional) Some instructors expect their students to expand the denominator.
$$ \begin{align*} \frac{- 6x - 45}{(x + 6)(x+5)} = \frac{- 6x - 45}{x^2 + 11x + 30} \end{align*} $$
Answer
$$ \begin{align*} \frac{x - 10 }{x + 5} + \frac{x - 3}{x + 6} = \frac{- 6x - 45}{(x + 6)(x+5)} \end{align*} $$
Problem 7
Step 1
Identify the common denominator.
In this case the common denominator is $$(x+2)(2x-5)$$.
Step 2
Rewrite the expressions so they have the common denominator.
$$ \begin{align*} \frac{4x}{x + 2} - \frac{3x+1}{2x - 5} & = \blue{\frac{2x-5}{2x-5}}\cdot\frac{4x}{x + 2} - \frac{3x+1}{2x - 5}\cdot\red{\frac{x+2}{x+2}}\\[6pt] & = \frac{4x\blue{(2x-5)}}{\blue{(2x-5)}(x + 2)} - \frac{(3x+1)\red{(x+2)}}{(2x - 5)\red{(x+2)}} \end{align*} $$
Step 3
Expand the numerators.
$$ \begin{align*} \frac{4x\blue{(2x-5)}}{\blue{(2x-5)}(x + 2)} - \frac{(3x+1)\red{(x+2)}}{(2x - 5)\red{(x+2)}} & = \frac{\blue{8x^2-20x}}{(2x-5)(x + 2)} - \frac{\red{3x^2+7x+2}}{(2x - 5)(x+2)} \end{align*} $$
Step 4
Subtract the numerators, but keep the same denominator. Be sure to sure to subtract the entire second denominator!
$$ \begin{align*} \frac{\blue{8x^2-20x}}{(2x-5)(x + 2)} - \frac{\red{3x^2+7x+2}}{(2x - 5)(x+2)} & = \frac{\blue{8x^2-20x} - (\red{3x^2+7x+2})}{(2x - 5)(x+2)}\\[6pt] & = \frac{\blue{8x^2-20x} - \red{3x^2 - 7x - 2}}{(2x - 5)(x+2)}\\[6pt] & = \frac{\blue{8x^2} - \red{3x^2} - \blue{20x} - \red{7x} - \red{2}}{(2x - 5)(x+2)}\\[6pt] & = \frac{5x^2 - 27x - 2}{(2x - 5)(x+2)} \end{align*} $$
Step 5
(Optional) Some instructors require their students to expand the denominator.
$$ \begin{align*} \frac{5x^2 - 27x - 2}{(2x - 5)(x+2)} = \frac{5x^2 - 27x - 2}{2x^2 - x - 10} \end{align*} $$
Answer
$$ \begin{align*} \frac{4x}{x + 2} - \frac{3x+1}{2x - 5} = \frac{5x^2 - 27x - 2}{(2x - 5)(x+2)} \end{align*} $$
Problem 8
Step 1
Identify the common denominator.
In this case, the common denominator is $$(3x - 5)(7x+4)$$
Step 2
Rewrite the expressions so they have the common denominator.
$$ \begin{align*} \frac{5x+2}{3x - 5} + \frac{3x+1}{7x + 4} & = \blue{\frac{7x+4}{7x+4}}\cdot\frac{5x+2}{3x - 5} + \frac{3x+1}{7x + 4}\cdot \red{\frac{3x-5}{3x-5}}\\[6pt] & = \frac{\blue{(7x+4)}(5x+2)}{\blue{(7x+4)}(3x - 5)} + \frac{(3x+1)\red{(3x-5)}}{(7x + 4)\red{(3x-5)}} \end{align*} $$
Step 3
Expand the numerators.
$$ \begin{align*} \frac{\blue{(7x+4)}(5x+2)}{\blue{(7x+4)}(3x - 5)} + \frac{(3x+1)\red{(3x-5)}}{(7x + 4)\red{(3x-5)}} & = \frac{\blue{35x^2 + 34x + 8}}{(7x+4)(3x - 5)} + \frac{\red{9x^2-12x-5}}{(7x + 4)(3x-5)} \end{align*} $$
Step 4
Add the numerators, but keep the same denominator.
$$ \begin{align*} \frac{\blue{35x^2 + 34x + 8}}{(7x+4)(3x - 5)} + \frac{\red{9x^2-12x-5}}{(7x + 4)(3x-5)} & = \frac{\blue{35x^2 + 34x + 8} + \red{9x^2-12x-5}}{(7x + 4)(3x-5)}\\[6pt] & = \frac{\blue{35x^2} + \red{9x^2} + \blue{34x} - \red{12x} + \blue{8} - \red{5}}{(7x + 4)(3x-5)}\\[6pt] & = \frac{44x^2 + 22x + 3}{(7x + 4)(3x-5)} \end{align*} $$
Step 5
(Optional) Some instructors require their students to expand the denominator.
$$ \begin{align*} \frac{44x^2 + 22x + 3}{(7x + 4)(3x-5)} = \frac{44x^2 + 22x + 3}{21x^2 -23x - 20} \end{align*} $$
Answer
$$ \begin{align*} \frac{5x+2}{3x - 5} + \frac{3x+1}{7x + 4} = \frac{44x^2 + 22x + 3}{(7x + 4)(3x-5)} \end{align*} $$
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