Adding and Subtracting with Common Denominators
Adding and subtracting rational expressions is similar to adding and subtracting fractions. Recall that if the denominators are the same, we can add or subtract the numerators and write the result over the common denominator.
\(\begin{aligned} \frac{3}{13}+\frac{7}{13} &=\frac{3+7}{13} \\ &=\frac{10}{13} \end{aligned}\)
When working with rational expressions, the common denominator will be a polynomial. In general, given polynomials P, Q, and R, where \(Q≠0\), we have the following:
\[\frac{P}{Q} \pm \frac{R}{Q}=\frac{P \pm R}{Q}\]
In this section, assume that all variable factors in the denominator are nonzero.
Example \(\PageIndex{1}\)
Add:
\(\frac{3}{y}+\frac{7}{y}\)
Solution:
Add the numerators \(3\) and \(7\), and write the result over the common denominator, y.
\(\begin{aligned} \frac{3}{y}+\frac{7}{y} &=\frac{3+7}{y} \\ &=\frac{10}{y} \end{aligned}\)
Answer:
\(\frac{10}{y}\)
Example \(\PageIndex{2}\)
Subtract:
\(\frac{x-5}{2 x-1}-\frac{1}{2 x-1}\)
Solution:
Subtract the numerators \(x−5\) and \(1\), and write the result over the common denominator, \(2x−1\).
\(\begin{aligned} \frac{x-5}{2 x-1}-\frac{1}{2 x-1} &=\frac{x-5-1}{2 x-1}\qquad\color{Cerulean}{Simplify\:the\:numerator.} \\ &=\frac{x-6}{2 x-1} \end{aligned}\)
Answer:
\(\frac{x-6}{2 x-1}\)
Example \(\PageIndex{3}\)
Subtract:
\(\frac{2 x+7}{(x+5)(x-3)}-\frac{x+10}{(x+5)(x-3)}\)
Solution:
We use parentheses to remind us to subtract the entire numerator of the second rational expression.
\(\begin{aligned} \frac{2 x+7}{(x+5)(x-3)}-\frac{x+10}{(x+5)(x-3)} &=\frac{(2 x+7)-(x+10)}{(x+5)(x-3)}\qquad\color{Cerulean}{Simplify\:the\:numerator.} \\ &=\frac{2 x+7-x-10}{(x+5)(x-3)}\qquad\quad\:\:\color{Cerulean}{Leave\:the\:denominator\:factored.} \\ &=\frac{\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{x-3}}}}}{(x+5)\color{Cerulean}{\cancel{\color{black}{(x-3)}}}}\qquad\:\:\:\quad\color{Cerulean}{Cancel\:common\:factors.} \\ &=\frac{1}{x+5} \end{aligned}\)
Answer:
\(\frac{1}{x+5}\)
Example \(\PageIndex{4}\)
Simplify:
\(\frac{2 x^{2}+10 x+3}{x^{2}-36}-\frac{x^{2}+6 x+5}{x^{2}-36}+\frac{x-4}{x^{2}-36}\)
Solution:
Subtract and add the numerators. Make use of parentheses and write the result over the common denominator, \(x^{2}−36\).
Answer:
\(\frac{x-1}{x-6}\)
Exercise \(\PageIndex{1}\)
Subtract:
- Answer
-
\(\frac{1}{x-4}\)
Adding and Subtracting with Unlike Denominators
To add rational expressions with unlike denominators, first find equivalent expressions with common denominators. Do this just as you have with fractions. If the denominators of fractions are relatively prime, then the least common denominator (LCD) is their product. For example,
\(\frac{1}{3}+\frac{1}{5} \color{Cerulean}{\Rightarrow}\color{black}{ \mathrm{LCD}=3 \cdot 5=15}\)
Multiply each fraction by the appropriate form of 1 to obtain equivalent fractions with a common denominator.
\(\begin{aligned} \frac{1}{3}+\frac{1}{5} &=\frac{1\color{Cerulean}{ \cdot 5}}{3 \color{Cerulean}{\cdot 5}}+\frac{1 \color{Cerulean}{\cdot 3}}{5 \color{Cerulean}{\cdot 3}} \\ &=\frac{5}{15}+\frac{3}{15}\qquad\qquad\color{Cerulean}{Equivalent\:fractions\:with\:a\:common\:denominator} \\ &=\frac{5+3}{15} \\ &=\frac{8}{15} \end{aligned}\)
The process of adding and subtracting rational expressions is similar. In general, given polynomials P, Q, R, and S, where \(Q≠0\) and \(S≠0\), we have the following:
\[\frac{P}{Q} \pm \frac{R}{S}=\frac{P S \pm Q R}{Q S}\]
In this section, assume that all variable factors in the denominator are nonzero.
Example \(\PageIndex{5}\)
Add:
\(\frac{1}{x}+\frac{1}{y}\)
Solution:
In this example, the \(LCD=xy\). To obtain equivalent terms with this common denominator, multiply the first term by \(\frac{y}{y}\) and the second term by \(\frac{x}{x}\).
\(\begin{aligned} \frac{1}{x}+\frac{1}{y} &=\frac{1}{x} \cdot\color{Cerulean}{ \frac{y}{y}}\color{black}{+}\frac{1}{y} \cdot \color{Cerulean}{\frac{x}{x}} \\ &=\frac{y}{x y}+\frac{x}{x y}\qquad\qquad\color{Cerulean}{Equivalent\:terms\:with\:a\:common\:denominator.} \\ &=\frac{y+x}{x y}\qquad\qquad\:\:\quad\color{Cerulean}{Add\:the\:numerators\:and\:place\:the\:result\:over\:the\:common\:denominator,\:xy.} \end{aligned}\)
Answer:
\(\frac{y+x}{x y}\)
Example \(\PageIndex{6}\)
Subtract:
\(\frac{1}{y}-\frac{1}{y-3}\)
Solution:
Since the \(LCD=y(y−3)\), multiply the first term by 1 in the form of \(\frac{(y−3)}{(y−3)}\) and the second term by \(\frac{y}{y}\).
\(\begin{aligned} \frac{1}{y}-\frac{1}{y-3} &=\frac{1}{y} \cdot\color{Cerulean}{ \frac{(y-3)}{(y-3)}}\color{black}{-}\frac{1}{y-3} \cdot\color{Cerulean}{ \frac{y}{y}} \\ &=\frac{(y-3)}{y(y-3)}-\frac{y}{y(y-3)} \\ &=\frac{y-3-y}{y(y-3)} \\ &=\frac{-3}{y(y-3)} \quad \text { or }=-\frac{3}{y(y-3)} \end{aligned}\)
Answer:
\(\frac{-3}{y(y-3)}\)
It is not always the case that the LCD is the product of the given denominators. Typically, the denominators are not relatively prime; thus determining the LCD requires some thought. Begin by factoring all denominators. The LCD is the product of all factors with the highest power. For example, given
\(\frac{1}{\color{Cerulean}{x^{3}}\color{black}{(x+2)}\color{Cerulean}{(x-3)}} \quad \text { and } \quad \frac{1}{x\color{Cerulean}{(x+2)^{2}}}\)
there are three base factors in the denominator: \(x, (x+2)\), and \((x−3)\). The highest powers of these factors are \(x^{3}, (x+2)^{2}\), and \((x−3)^{1}\). Therefore,
\(\mathrm{LCD}=\color{Cerulean}{x^{3}(x+2)^{2}(x-3)}\)
The general steps for adding or subtracting rational expressions are illustrated in the following example.
Example \(\PageIndex{7}\)
Subtract:
Solution:
Step 1: Factor all denominators to determine the LCD.
The LCD is \((x+1)(x+3)(x−5)\).
Step 2: Multiply by the appropriate factors to obtain equivalent terms with a common denominator. To do this, multiply the first term by \(\frac{(x−5)}{(x−5)}\) and the second term by \(\frac{(x+3)}{(x+3)}\).
\(\begin{array}{l}{=\frac{x}{(x+1)(x+3)} \cdot \color{Cerulean}{\frac{(x-5)}{(x-5)}}\color{black}{-}\frac{3}{(x+1)(x-5)} \cdot\color{Cerulean}{ \frac{(x+3)}{(x+3)}}} \\ {=\frac{x(x-5)}{(x+1)(x+3)(x-5)}-\frac{3(x+3)}{(x+1)(x+3)(x-5)}}\end{array}\)
Step 3: Add or subtract the numerators and place the result over the common denominator.
\(=\frac{x(x-5)-3(x+3)}{(x+1)(x+3)(x-5)}\)
Step 4: Simplify the resulting algebraic fraction.
Answer:
\(\frac{(x-9)}{(x+3)(x-5)}\)
Example \(\PageIndex{8}\)
Subtract:
Solution:
It is best not to factor the numerator, \(x^{2}−9x+18\), because we will most likely need to simplify after we subtract.
Answer:
\(\frac{18}{(x-4)(x-9)}\)
Example \(\PageIndex{9}\)
Subtract:
\(\frac{1}{x^{2}-4}-\frac{1}{2-x}\)
Solution:
First, factor the denominators and determine the LCD. Notice how the opposite binomial property is applied to obtain a more workable denominator.
\(\begin{aligned} \frac{1}{x^{2}-4}-\frac{1}{2-x} &=\frac{1}{(x+2)(x-2)}-\frac{1}{-1(x-2)} \\ &=\frac{1}{(x+2)(x-2)}+\frac{1}{(x-2)} \end{aligned}\)
The LCD is \((x+2)(x−2)\). Multiply the second term by 1 in the form of \(\frac{(x+2)}{(x+2)}\).
\(\begin{array}{l}{=\frac{1}{(x+2)(x-2)}+\frac{1}{(x-2)} \cdot \color{Cerulean}{\frac{(x+2)}{(x+2)}}} \\ {=\frac{1}{(x+2)(x-2)}+\frac{x+2}{(x-2)(x+2)}}\end{array}\)
Now that we have equivalent terms with a common denominator, add the numerators and write the result over the common denominator.
\(\begin{array}{l}{=\frac{1}{(x+2)(x-2)}+\frac{x+2}{(x-2)(x+2)}} \\ {=\frac{1+x+2}{(x+2)(x-2)}} \\ {=\frac{x+3}{(x+2)(x-2)}}\end{array}\)
Answer:
\(\frac{x+3}{(x+2)(x-2)}\)
Example \(\PageIndex{10}\)
Simplify:
\(\frac{y-1}{y+1}-\frac{y+1}{y-1}+\frac{y^{2}-5}{y^{2}-1}\)
Solution:
Begin by factoring the denominator.
\(\begin{array}{c}{\frac{y-1}{y+1}-\frac{y+1}{y-1}+\frac{y^{2}-5}{y^{2}-1}} \\ {=\frac{y-1}{y+1}-\frac{y+1}{y-1}+\frac{y^{2}-5}{(y+1)(y-1)}}\end{array}\)
We can see that the LCD is \((y+1)(y−1)\). Find equivalent fractions with this denominator.
Next, subtract and add the numerators and place the result over the common denominator.
Finish by simplifying the resulting rational expression.
Answer:
\(\frac{y-5}{y-1}\)
Exercise \(\PageIndex{2}\)
Simplify:
\(-\frac{2 x^{2}-1+x}{1+x}-\frac{5}{1-x}\)
- Answer
-
\(\frac{x+3}{x-1}\)
Rational expressions are sometimes expressed using negative exponents. In this case, apply the rules for negative exponents before simplifying the expression.
Example \(\PageIndex{11}\)
Simplify:
\(y^{-2}+(y-1)^{-1}\)
Solution:
Recall that \(x^{−n}=\frac{1}{x^{n}}\). We begin by rewriting the negative exponents as rational expressions.
\(\begin{aligned} & y^{-2}+(y-1)^{-1} \\=& \frac{1}{y^{2}}+\frac{1}{(y-1)^{1}}\qquad\qquad\qquad\qquad\color{Cerulean}{Replace\:negative\:exponents.} \\=& \frac{1}{y^{2}} \cdot \color{Cerulean}{\frac{(y-1)}{(y-1)}}\color{black}{+\frac{1}{(y-1)}} \cdot\color{Cerulean}{ \frac{y^{2}}{y^{2}}}\qquad\color{Cerulean}{Multiply\:by\:factors\:to\:obtain\:equivalent}\\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{Cerulean}{expressions\:with\:a\:common\:denominator.} \\=&\frac{(y-1)}{y^{2}(y-1)}+\frac{y^{2}}{y^{2}(y-1)} \\=&\frac{(y-1)+y^{2}}{y^{2}(y-1)}\qquad\qquad\qquad\qquad\:\:\:\color{Cerulean}{Add\:and\:simplify.} \\ =& \frac{y^{2}+y-1}{y^{2}(y-1)}\qquad\qquad\qquad\qquad\quad\:\:\:\color{Cerulean}{The\:trinomial\:does\:not\:factor.} \end{aligned}\)
Answer:
\(\frac{y^{2}+y-1}{y^{2}(y-1)}\)
