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Adding Rational Expressions With a Common Denominator

Adding Rational Expressions With a Common Denominator

What is the first step you take when you add numerical fractions? You check if they have a common denominator. If they do, you add the numerators and place the sum over the common denominator. If they do not have a common denominator, you find one before you add.

It is the same with rational expressions. To add rational expressions, they must have a common denominator. When the denominators are the same, you add the numerators and place the sum over the common denominator.

Rational Expression Addition

If \(p,q,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r\) are polynomials where \(r\ne 0\), then

\(\frac{p}{r}+\frac{q}{r}=\frac{p+q}{r}\)

To add rational expressions with a common denominator, add the numerators and place the sum over the common denominator.

We will add two numerical fractions first, to remind us of how this is done.

Example

Add: \(\frac{5}{18}+\frac{7}{18}.\)

Solution

\(\begin{array}{cccc}& & & \hfill \phantom{\rule{5em}{0ex}}\frac{5}{18}+\frac{7}{18}\hfill \\ \\ \begin{array}{c}\text{The fractions have a common}\hfill \\ \text{denominator, so add the numerators and}\hfill \\ \text{place the sum over the common}\hfill \\ \text{denominator.}\hfill \end{array}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\frac{5+7}{18}\hfill \\ \\ \text{Add in the numerator.}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\frac{12}{18}\hfill \\ \\ \begin{array}{c}\text{Factor the numerator and denominator to}\hfill \\ \text{show the common factors.}\hfill \end{array}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\frac{6·2}{6·3}\hfill \\ \\ \text{Remove common factors.}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\frac{\require{cancel}\cancel{6}·2}{\require{cancel}\cancel{6}·3}\hfill \\ \\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\frac{2}{3}\hfill \end{array}\)

Remember, we do not allow values that would make the denominator zero. What value of \(y\) should be excluded in the next example?

Example

Add: \(\frac{3y}{4y-3}+\frac{7}{4y-3}.\)

Solution

\(\begin{array}{cccc}& & & \hfill \phantom{\rule{5em}{0ex}}\frac{3y}{4y-3}+\frac{7}{4y-3}\hfill \\ \begin{array}{c}\text{The fractions have a common}\hfill \\ \text{denominator, so add the numerators and}\hfill \\ \text{place the sum over the common}\hfill \\ \text{denominator.}\hfill \end{array}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\frac{3y+7}{4y-3}\hfill \end{array}\)

The numerator and denominator cannot be factored. The fraction is simplified.

Example

Add: \(\frac{7x+12}{x+3}+\frac{{x}^{2}}{x+3}.\)

Solution

\(\begin{array}{cccc}& & & \hfill \phantom{\rule{5em}{0ex}}\frac{7x+12}{x+3}+\frac{{x}^{2}}{x+3}\hfill \\ \\ \begin{array}{c}\text{The fractions have a common}\hfill \\ \text{denominator, so add the numerators and}\hfill \\ \text{place the sum over the common}\hfill \\ \text{denominator.}\hfill \end{array}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\frac{7x+12+{x}^{2}}{x+3}\hfill \\ \\ \text{Write the degrees in descending order.}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\frac{{x}^{2}+7x+12}{x+3}\hfill \\ \\ \text{Factor the numerator.}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\frac{\left(x+3\right)\left(x+4\right)}{x+3}\hfill \\ \\ \text{Simplify by removing common factors.}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\frac{\require{cancel}\cancel{\left(x+3\right)}\left(x+4\right)}{\require{cancel}\cancel{x+3}}\hfill \\ \\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}x+4\hfill \end{array}\)

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