8.4: Add and Subtract Rational Expressions
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Adding and subtracting rational expressions are identical to adding and subtracting with numerical fractions. Recall, when adding with a common denominator, we add across numerators and keep the same denominator. This is the same method we use with rational expressions. Note, methods never change, only problems.
For adding and subtracting with rational expressions, here are some helpful tips:
- Identify the denominators: are they the same or different?
- Combine the rational expressions into one expression.
- Once combined into one expression, then reduce the fraction, if possible.
- A fraction is reducible only if there is a gcf between the numerator and denominator.
- If the numerator and denominator cannot be factored, they are unlikely to have any common factors.
Add or Subtract Rational Expressions with a Common Denominator
Recall. We can use the same properties for adding or subtracting fractions with common denominators also for adding and subtracting rational expressions with common denominators:
\[\dfrac{a}{c}\pm\dfrac{b}{c}=\dfrac{a\pm b}{c}\nonumber\]
Add: \(\dfrac{x-4}{x^2-2x-8}+\dfrac{x+8}{x^2-2x-8}\)
Solution
Using the helpful tips above, the denominators are the same. Let’s combine into one fraction by adding across numerators and keeping the denominator the same:
\[\begin{array}{rl}\dfrac{x-4}{x^2-2x-8}+\dfrac{x+8}{x^2-2x-8}&\text{Like denominators, add across numerators} \\ \dfrac{2x+4}{x^2-2x-8}&\text{Factor the numerator and denominator} \\ \dfrac{2(x+2)}{(x+2)(x-4)}&\text{Reduce out a factor of }(x+2) \\ \dfrac{2\color{blue}{\cancel{(x+2)}}}{\color{blue}{\cancel{(x+2)}}\color{black}{}(x-4)}&\text{Rewrite} \\ \dfrac{2}{(x-4)}&\text{Sum}\end{array}\nonumber\]
Notice, we had a GCF in the numerator. This is when we know the fraction may be reducible and we factor the GCF and determine whether the expression is reducible.
Subtraction with common denominators follows the same pattern. However, with subtraction, we first distribute the subtraction through the numerator. Then simplify as usual. This process is the same as “add the opposite” when subtracting with negative integers.
Subtract: \(\dfrac{6x-12}{3x-6}-\dfrac{15x-6}{3x-6}\)
Solution
Using the helpful tips above, the denominators are the same. Let’s combine into one fraction by subtracting across numerators and keeping the denominator the same:
\[\begin{array}{rl}\dfrac{6x-12}{3x-6}-\dfrac{15x-6}{3x-6}&\text{Like denominators, subtract across numerators} \\ \dfrac{6x-12\color{blue}{-(15x-6)}}{3x-6}&\color{black}{\text{Simplify the numerator}} \\ \dfrac{-9x-6}{3x-6}&\text{Factor the numerator and denominator} \\ \dfrac{-3(3x+2)}{3(x-2)}&\text{Reduce out a factor of }3 \\ \dfrac{-\color{blue}{\cancel{3}}\color{black}{}(3x+2)}{\color{blue}{\cancel{3}}\color{black}{}(x-2)}&\text{Rewrite} \\ \dfrac{-(3x+2)}{x-2}&\text{Difference}\end{array}\nonumber\]
Notice, we had a GCF in the numerator. This is when we know the fraction may be reducible and we factor the GCF and determine whether the expression is reducible.
The Rhind papyrus of Egypt from 1650 BC gives some of the earliest known symbols for addition and subtraction. For addition, a pair of legs walking in the direction one reads, and for subtraction, a pair of legs walking in the opposite direction.
Add and Subtract Rational Expressions with Unlike Denominators
Recall. We can use the same properties for adding and subtracting integer fractions with unlike denominators for adding and subtracting rational expressions with unlike denominators.
Add: \(\dfrac{7a}{3a^2b}+\dfrac{4b}{6ab^4}\)
Solution
Using the helpful tips above, the denominators are different. We need to find the LCD, rewrite each fraction with the LCD, then combine into one fraction.
\[\begin{array}{rl}\dfrac{7a}{3a^2b}+\dfrac{4b}{6ab^4}&\text{Unlike denominators; LCD}=6a^2b^4 \\ \color{blue}{\dfrac{2b^3}{2b^3}}\color{black}{}\cdot\dfrac{7a}{3a^2b}+\dfrac{4b}{6ab^4}\cdot\color{blue}{\dfrac{a}{a}}&\color{black}{\text{Rewrite each fraction with the LCD}} \\ \dfrac{14ab^3}{6a^2b^4}+\dfrac{4ab}{6a^2b^4}&\text{Same denominator, add across numerators} \\ \dfrac{14ab^3+4ab}{6a^2b^4}&\text{Factor the numerator}\\ \dfrac{2ab(7b^3+2)}{6a^2b^4}&\text{Reduce out a factor of }2ab \\ \dfrac{\color{blue}{\cancel{2ab}}\color{black}{}(7b^3+2)}{\color{blue}{\cancelto{3}{6}}\color{black}{}a^{\color{blue}{\cancelto{1}{2}}}\color{black}{}b^{\color{blue}{\cancelto{3}{4}}} }&\color{black}{\text{Rewrite}} \\ \dfrac{7b^3+2}{3ab^3}&\text{Sum}\end{array}\nonumber\]
Since there isn’t a GCF in the numerator, as stated in the helpful tips, we cannot further reduce the fraction.
Subtract: \(\dfrac{4}{5a}-\dfrac{7b}{4a^2}\)
Solution
Using the helpful tips above, the denominators are different. We need to find the LCD, rewrite each fraction with the LCD, then combine into one fraction.
\[\begin{array}{rl}\dfrac{4}{5a}-\dfrac{7b}{4a^2}&\text{Unlike denominators; LCD}=20a^2 \\ \color{blue}{\dfrac{4a}{4a}}\color{black}{}\cdot\dfrac{4}{5a}-\dfrac{7b}{4a^2}\cdot\color{blue}{\dfrac{5}{5}}&\color{black}{\text{Rewrite each fraction with the LCD}} \\ \dfrac{16a}{20a^2}-\dfrac{35b}{20a^2}&\text{Same denominator, subtract across numerators} \\ \dfrac{16a-35b}{20a^2}&\text{Difference}\end{array}\nonumber\]
Since there isn’t a GCF in the numerator, as stated in the helpful tips, we cannot further reduce the fraction.
Add: \(\dfrac{6}{8a+4}+\dfrac{3a}{8}\)
Solution
Using the helpful tips above, the denominators are different. We need to find the LCD, rewrite each fraction with the LCD, then combine into one fraction. To determine the LCD, we have to factor the binomial in the first fraction’s denominator.
\[\begin{array}{rl}\dfrac{6}{8a+4}+\dfrac{3a}{8}&\text{Factor the first denominator} \\ \dfrac{6}{4(2a+1)}+\dfrac{3a}{8}&\text{Unlike denominators; LCD}=8(2a+1) \\ \color{blue}{\dfrac{2}{2}}\color{black}{}\cdot\dfrac{6}{4(2a+1)}+\dfrac{3a}{8}\cdot\color{blue}{\dfrac{(2a+1)}{(2a+1)}}&\color{black}{\text{Rewrite each fraction with the LCD}} \\ \dfrac{12}{8(2a+1)}+\dfrac{3a(2a+1)}{8(2a+1)}&\text{Same denominator, add across numerators} \\ \dfrac{12+6a^2+3a}{8(2a+1)}&\text{Factor the numerator} \\ \dfrac{3(2a^2+a+4)}{8(2a+1)}&\text{Expression is irreducible} \\ \dfrac{3(2a^2+a+4)}{8(2a+1)}&\text{Sum}\end{array}\nonumber\]
Notice there is a GCF of \(3\) in the numerator, but \(3\) isn’t a common factor. However, we still need to factor the GCF where possible from the numerator to verify that we can either reduce or not reduce the expression.
Subtract: \(\dfrac{x+1}{x-4}-\dfrac{x+1}{x^2-7x+12}\)
Solution
Using the helpful tips above, the denominators are different. We need to find the LCD, rewrite each fraction with the LCD, then combine into one fraction. To determine the LCD, we have to factor the trinomial in the second fraction's denominator.
\[\begin{array}{rl}\dfrac{x+1}{x-4}-\dfrac{x+1}{x^2-7x+12}&\text{Factor the second denominator} \\ \dfrac{x+1}{(x-4)}-\dfrac{x+1}{(x-3)(x-4)}&\text{Unlike denominators; LCD}=(x-3)(x-4) \\ \color{blue}{\dfrac{(x-3)}{(x-3)}}\color{black}{}\cdot\dfrac{(x+1)}{(x-4)}-\dfrac{(x+1)}{(x-3)(x-4)}&\text{Rewrite each fraction with the LCD} \\ \dfrac{(x-3)(x+1)}{(x-3)(x-4)}-\dfrac{(x+1)}{(x-3)(x-4)}&\text{Same denominator, FOIL, subtract across numerators} \\ \dfrac{x^2-2x-3\color{blue}{-(x-1)}}{(x-3)(x-4)}&\text{Simplify the numerator} \\ \dfrac{x^2-3x-4}{(x-3)(x-4)}&\text{Factor the numerator} \\ \dfrac{(x-4)(x+1)}{(x-3)(x-4)}&\text{Reduce out a factor of }(x-4) \\ \dfrac{\color{blue}{\cancel{(x-4)}}\color{black}{}(x+1)}{(x-3)\color{blue}{\cancel{(x-4)}}}&\color{black}{\text{Rewrite}} \\ \dfrac{(x+1)}{(x-3)}&\text{Difference}\end{array}\nonumber\]
Recall, we do not reduce terms, only factors. Thus, the fraction above is the difference.
We are not allowed to reduce terms, only factors.
Add and Subtract with Rational Functions
Let \(P(x)=\dfrac{x+6}{x+5}\) and \(R(x)=\dfrac{x+3}{x-9}\). Add and simplify \((P+R)(x)\).
Solution
Using the helpful tips above, the denominators are different. We need to find the LCD, rewrite each fraction with the LCD, then combine into one fraction. To determine the LCD, we look at the denominators.
\[\begin{array}{rl}(P+R)(x)=P(x)+R(x)&\text{Replace }P\text{ and }R \\ (P+R)(x)=\dfrac{x+6}{x+5}+\dfrac{x+3}{x-9}&\text{Unlike denominators; LCD}=(x+5)(x-9) \\ (P+R)(x)=\color{blue}{\dfrac{(x-9)}{(x-9)}}\color{black}{}\cdot\dfrac{(x+6)}{(x+5)}+\dfrac{(x+3)}{(x-9)}\cdot\color{blue}{\dfrac{(x+5)}{(x+5)}}&\color{black}{\text{Rewrite each fraction with the LCD}} \\ (P+R)(x)=\dfrac{(x-9)(x+6)}{(x-9)(x+5)}+\dfrac{(x+3)(x+5)}{(x-9)(x+5)}&\text{Multiply each numerator} \\ (P+R)(x)=\dfrac{x^2-3x-54}{(x-9)(x+5)}+\dfrac{x^2+8x+15}{(x-9)(x+5)}&\text{Same denominator, add across numerators} \\ (P+R)(x)=\dfrac{2x^2+5x-39}{(x-9)(x+5)}&\text{Expression is irreducible} \\ (P+R)(x)=\dfrac{2x^2+5x-39}{(x-9)(x+5)}&\text{Sum of }P\text{ and }R\end{array}\nonumber\]
Since the numerator isn’t factorable, as stated in the helpful tips, we cannot further reduce the fraction.
Let \(f(x)=\dfrac{x-3}{x+5}\) and \(g(x)=\dfrac{-5x+7}{x^2+6x+5}\). Subtract and simplify \((f-g)(x)\).
Solution
Using the helpful tips above, the denominators are different. We need to find the LCD, rewrite each fraction with the LCD, then combine into one fraction. To determine the LCD, we look at the denominators.
\[\begin{array}{rl}(f-g)(x)=f(x)-g(x)&\text{Replace }f\text{ and }g \\ (f-g)(x)=\dfrac{x-3}{x+5}-\dfrac{-5x+7}{x^2+6x+5}&\text{Factor the second denominator} \\ (f-g)(x)=\dfrac{x-3}{x+5}-\dfrac{-5x+7}{(x+5)(x+1)}&\text{Unlike denominators; LCD}=(x+5)(x+1) \\ (f-g)(x)=\color{blue}{\dfrac{(x+1)}{(x+1)}}\color{black}{}\cdot\dfrac{(x-3)}{(x+5)}-\dfrac{-5x+7}{(x+5)(x+1)}&\text{Rewrite each fraction with the LCD} \\ (f-g)(x)=\dfrac{(x+1)(x-3)}{(x+5)(x+1)}-\dfrac{-5x+7}{(x+5)(x+1)}&\text{Multiply the first numerator} \\ (f-g)(x)=\dfrac{x^2-2x-3}{(x+5)(x+1)}-\dfrac{-5x+7}{(x+5)(x+1)}&\text{Same denominator, subtract across numerators} \\ (f-g)(x)=\dfrac{x^2-2x-3-(-5x+7)}{(x+5)(x+1)}&\text{Simplify the numerator}\\ (f-g)(x)=\dfrac{x^2+3x-10}{(x+5)(x+1)}&\text{Factor the numerator} \\ (f-g)(x)=\dfrac{(x+5)(x-2)}{(x+5)(x+1)}&\text{Reduce out a factor of }x+5 \\ (f-g)(x)=\dfrac{\color{blue}{\cancel{(x+5)}}\color{black}{}(x-2)}{\color{blue}{\cancel{(x+5)}}\color{black}{}(x+1)}&\text{Rewrite} \\ (f-g)(x)=\dfrac{(x-2)}{(x+1)}&\text{Difference of }f\text{ and }g\end{array}\nonumber\]
Add and Subtract Rational Expressions Homework
Add or subtract the rational expressions. Simplify completely.
\(\dfrac{2}{a+3}+\dfrac{4}{a+3}\)
\(\dfrac{t^2+4t}{t-1}+\dfrac{2t-7}{t-1}\)
\(\dfrac{2x^2+3}{x^2-6x+5}-\dfrac{x^2-5x+9}{x^2-6x+5}\)
\(\dfrac{5}{6r}-\dfrac{5}{8r}\)
\(\dfrac{8}{9t^3}+\dfrac{5}{6t^2}\)
\(\dfrac{a+2}{2}-\dfrac{a-4}{4}\)
\(\dfrac{x-1}{4x}-\dfrac{2x+3}{x}\)
\(\dfrac{5x+3y}{2x^2y}-\dfrac{3x+4y}{xy^2}\)
\(\dfrac{2z}{z-1}-\dfrac{3z}{z+1}\)
\(\dfrac{8}{x^2-4}-\dfrac{3}{x+2}\)
\(\dfrac{t}{t-3}-\dfrac{5}{4t-12}\)
\(\dfrac{2}{5x^2+5x}-\dfrac{4}{3x+3}\)
\(\dfrac{t}{y-t}-\dfrac{y}{y+t}\)
\(\dfrac{x}{x^2+5x+6}-\dfrac{2}{x^2+3x+2}\)
\(\dfrac{x}{x^2+15x+56}-\dfrac{7}{x^2+13x+42}\)
\(\dfrac{5x}{x^2-x-6}-\dfrac{18}{x^2-9}\)
\(\dfrac{2x}{x^2-1}-\dfrac{4}{x^2+2x-3}\)
\(\dfrac{x+1}{x^2-2x-35}+\dfrac{x+6}{x^2+7x+10}\)
\(\dfrac{4-a^2}{a^2-9}-\dfrac{a-2}{3-a}\)
\(\dfrac{2z}{1-2z}+\dfrac{3z}{2z+1}-\dfrac{3}{4z^2-1}\)
\(\dfrac{2x-3}{x^2+3x+2}+\dfrac{3x-1}{x^2+5x+6}\)
\(\dfrac{2x+7}{x^2-2x-3}-\dfrac{3x-2}{x^2+6x+5}\)
\(\dfrac{x^2}{x-2}-\dfrac{6x-8}{x-2}\)
\(\dfrac{a^2+3a}{a^2+5a-6}-\dfrac{4}{a^2+5a-6}\)
\(\dfrac{3}{x}+\dfrac{4}{x^2}\)
\(\dfrac{7}{xy^2}+\dfrac{3}{x^2y}\)
\(\dfrac{x+5}{8}+\dfrac{x-3}{12}\)
\(\dfrac{2a-1}{3a^2}+\dfrac{5a+1}{9a}\)
\(\dfrac{2c-d}{c^2d}-\dfrac{c+d}{cd^2}\)
\(\dfrac{2}{x-1}+\dfrac{2}{x+1}\)
\(\dfrac{2}{x-5}+\dfrac{3}{4x}\)
\(\dfrac{4x}{x^2-25}+\dfrac{x}{x+5}\)
\(\dfrac{2}{x+3}+\dfrac{4}{(x+3)^2}\)
\(\dfrac{3a}{4a-20}+\dfrac{9a}{6a-30}\)
\(\dfrac{x}{x-5}+\dfrac{x-5}{x}\)
\(\dfrac{2x}{x^2-1}-\dfrac{3}{x^2+5x+4}\)
\(\dfrac{2x}{x^2-9}+\dfrac{5}{x^2+x-6}\)
\(\dfrac{4x}{x^2-2x-3}-\dfrac{3}{x^2-5x+6}\)
\(\dfrac{x-1}{x^2+3x+2}+\dfrac{x+5}{x^2+4x+3}\)
\(\dfrac{3x+2}{3x+6}+\dfrac{x}{4-x^2}\)
\(\dfrac{4y}{y^2-1}-\dfrac{2}{y}-\dfrac{2}{y+1}\)
\(\dfrac{2r}{r^2-s^2}+\dfrac{1}{r+s}-\dfrac{1}{r-s}\)
\(\dfrac{x+2}{x^2-4x+3}+\dfrac{4x+5}{x^2+4x-5}\)
\(\dfrac{3x-8}{x^2+6x+8}+\dfrac{2x-3}{x^2+3x+2}\)
Perform the indicated operation and simplify.
Let \(P(x)=\dfrac{x}{x+6}\) and \(R(x)=\dfrac{5x+6}{x^2+8x+12}\). Add and simplify \((P+R)(x)\).
Let \(f(x)=\dfrac{x}{x+7}\) and \(g(x)=\dfrac{10x+42}{x^2+10x+21}\). Add and simplify \((f+g)(x)\).
Let \(S(x)=\dfrac{x-1}{x+4}\) and \(V(x)=\dfrac{1x+14}{x^2+6x+8}\). Subtract and simplify \((S-V)(x)\).
Let \(r(n)=\dfrac{n-2}{n-3}\) and \(q(n)=\dfrac{11n-25}{n^2+2n-15}\). Subtract and simplify \((r-q)(n)\).