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Solving Rational Equations

Solving Rational Equations 1.16 - Solution

arrow_back Return to Solving Rational Equations
We want to solve the given rational equation. 1145x+3=2312\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 44 is a factor of 12,12, and therefore the LCD is the product of (x+3)(x+3) and 12.12.
1145x+3=2312\dfrac{11}{4}-\dfrac{5}{x+3}=\dfrac{23}{12}
(1145x+3)(x+3)(12)=2312(x+3)(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)}}
114(x+3)(12)5x+3(x+3)(12)=2312(x+3)(12)\dfrac{11}{4}(x+3)(12)-\dfrac{5}{x+3}(x+3)(12)=\dfrac{23}{12}(x+3)(12)
114(x+3)(4)(3)5x+3(x+3)(12)=2312(x+3)(12)\dfrac{11}{4}(x+3)(4)(3)-\dfrac{5}{x+3}(x+3)(12)=\dfrac{23}{12}(x+3)(12)
114(x+3)(4)(3)5x+3(x+3)(12)=2312(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)11(x+3)(3)-5(12)=23(x+3)
Solve for xx
33(x+3)60=23(x+3)33(x+3)-60=23(x+3)
33x+9960=23(x+3)33x+99-60=23(x+3)
33x+39=23(x+3)33x+39=23(x+3)
33x+39=23x+6933x+39=23x+69
10x+39=6910x+39=69
10x=3010x=30
x=3x=3
Finally, we will check our result to make sure that it is not an extraneous solution.
1145x+3=?2312\dfrac{11}{4}-\dfrac{5}{x+3}\stackrel{?}{=}\dfrac{23}{12}
11453+3=?2312\dfrac{11}{4}-\dfrac{5}{{\color{#0000FF}{3}}+3}\stackrel{?}{=}\dfrac{23}{12}
Simplify left-hand side
11456=?2312\dfrac{11}{4}-\dfrac{5}{6}\stackrel{?}{=}\dfrac{23}{12}
331256=?2312\dfrac{33}{12}-\dfrac{5}{6}\stackrel{?}{=}\dfrac{23}{12}
33121012=?2312\dfrac{33}{12}-\dfrac{10}{12}\stackrel{?}{=}\dfrac{23}{12}
2312=2312 \dfrac{23}{12}=\dfrac{23}{12} \ {\color{#009600}{\huge{\checkmark}}}