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# Solving Rational Equations

## Solving Rational Equations 1.16 - Solution

We want to solve the given rational equation. $\begin{gathered} \dfrac{11}{4}-\dfrac{5}{x+3}=\dfrac{23}{12} \end{gathered}$ To do so, we will first determine the least common denominator (LCD) so that we can clear the denominators. Note that $4$ is a factor of $12,$ and therefore the LCD is the product of $(x+3)$ and $12.$
$\dfrac{11}{4}-\dfrac{5}{x+3}=\dfrac{23}{12}$
$\left(\dfrac{11}{4}-\dfrac{5}{x+3}\right){\color{#0000FF}{(x+3)}}{\color{#009600}{(12)}}=\dfrac{23}{12}{\color{#0000FF}{(x+3)}}{\color{#009600}{(12)}}$
$\dfrac{11}{4}(x+3)(12)-\dfrac{5}{x+3}(x+3)(12)=\dfrac{23}{12}(x+3)(12)$
$\dfrac{11}{4}(x+3)(4)(3)-\dfrac{5}{x+3}(x+3)(12)=\dfrac{23}{12}(x+3)(12)$
$\dfrac{11}{\cancel{{\color{#0000FF}{4}}}}(x+3)\cancel{{\color{#0000FF}{(4)}}}(3)-\dfrac{5}{\cancel{{\color{#009600}{x+3}}}}\cancel{{\color{#009600}{(x+3)}}}(12)=\dfrac{23}{\cancel{{\color{#FF0000}{12}}}}(x+3)\cancel{{\color{#FF0000}{(12)}}}$
$11(x+3)(3)-5(12)=23(x+3)$
Solve for $x$
$33(x+3)-60=23(x+3)$
$33x+99-60=23(x+3)$
$33x+39=23(x+3)$
$33x+39=23x+69$
$10x+39=69$
$10x=30$
$x=3$
Finally, we will check our result to make sure that it is not an extraneous solution.
$\dfrac{11}{4}-\dfrac{5}{x+3}\stackrel{?}{=}\dfrac{23}{12}$
$\dfrac{11}{4}-\dfrac{5}{{\color{#0000FF}{3}}+3}\stackrel{?}{=}\dfrac{23}{12}$
Simplify left-hand side
$\dfrac{11}{4}-\dfrac{5}{6}\stackrel{?}{=}\dfrac{23}{12}$
$\dfrac{33}{12}-\dfrac{5}{6}\stackrel{?}{=}\dfrac{23}{12}$
$\dfrac{33}{12}-\dfrac{10}{12}\stackrel{?}{=}\dfrac{23}{12}$
$\dfrac{23}{12}=\dfrac{23}{12} \ {\color{#009600}{\huge{\checkmark}}}$