We want to .
11 4 − 5 x + 3 = 23 12 \begin{gathered}
\dfrac{11}{4}-\dfrac{5}{x+3}=\dfrac{23}{12}
\end{gathered} 4 1 1 − x + 3 5 = 1 2 2 3
To do so, we will first determine the least (LCD) so that we can clear the denominators. Note that
4 4 4 is a factor of
12 , 12, 1 2 , and therefore the LCD is the product of
( x + 3 ) (x+3) ( x + 3 ) and
12. 12. 1 2 . 11 4 − 5 x + 3 = 23 12 \dfrac{11}{4}-\dfrac{5}{x+3}=\dfrac{23}{12} 4 1 1 − x + 3 5 = 1 2 2 3 ( 11 4 − 5 x + 3 ) ( x + 3 ) ( 12 ) = 23 12 ( 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)}} ( 4 1 1 − x + 3 5 ) ( x + 3 ) ( 1 2 ) = 1 2 2 3 ( x + 3 ) ( 1 2 ) 11 4 ( x + 3 ) ( 12 ) − 5 x + 3 ( x + 3 ) ( 12 ) = 23 12 ( x + 3 ) ( 12 ) \dfrac{11}{4}(x+3)(12)-\dfrac{5}{x+3}(x+3)(12)=\dfrac{23}{12}(x+3)(12) 4 1 1 ( x + 3 ) ( 1 2 ) − x + 3 5 ( x + 3 ) ( 1 2 ) = 1 2 2 3 ( x + 3 ) ( 1 2 ) 11 4 ( x + 3 ) ( 4 ) ( 3 ) − 5 x + 3 ( x + 3 ) ( 12 ) = 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) 4 1 1 ( x + 3 ) ( 4 ) ( 3 ) − x + 3 5 ( x + 3 ) ( 1 2 ) = 1 2 2 3 ( x + 3 ) ( 1 2 ) 11 4 ( x + 3 ) ( 4 ) ( 3 ) − 5 x + 3 ( x + 3 ) ( 12 ) = 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)}}} 4 1 1 ( x + 3 ) ( 4 ) ( 3 ) − x + 3 5 ( x + 3 ) ( 1 2 ) = 1 2 2 3 ( x + 3 ) ( 1 2 ) 11 ( x + 3 ) ( 3 ) − 5 ( 12 ) = 23 ( x + 3 ) 11(x+3)(3)-5(12)=23(x+3) 1 1 ( x + 3 ) ( 3 ) − 5 ( 1 2 ) = 2 3 ( x + 3 ) 33 ( x + 3 ) − 60 = 23 ( x + 3 ) 33(x+3)-60=23(x+3) 3 3 ( x + 3 ) − 6 0 = 2 3 ( x + 3 ) 33 x + 99 − 60 = 23 ( x + 3 ) 33x+99-60=23(x+3) 3 3 x + 9 9 − 6 0 = 2 3 ( x + 3 ) 33 x + 39 = 23 ( x + 3 ) 33x+39=23(x+3) 3 3 x + 3 9 = 2 3 ( x + 3 ) 33 x + 39 = 23 x + 69 33x+39=23x+69 3 3 x + 3 9 = 2 3 x + 6 9 10 x + 39 = 69 10x+39=69 1 0 x + 3 9 = 6 9
Finally, we will check our result to make sure that it is not an .
11 4 − 5 x + 3 = ? 23 12 \dfrac{11}{4}-\dfrac{5}{x+3}\stackrel{?}{=}\dfrac{23}{12} 4 1 1 − x + 3 5 = ? 1 2 2 3 11 4 − 5 3 + 3 = ? 23 12 \dfrac{11}{4}-\dfrac{5}{{\color{#0000FF}{3}}+3}\stackrel{?}{=}\dfrac{23}{12} 4 1 1 − 3 + 3 5 = ? 1 2 2 3 11 4 − 5 6 = ? 23 12 \dfrac{11}{4}-\dfrac{5}{6}\stackrel{?}{=}\dfrac{23}{12} 4 1 1 − 6 5 = ? 1 2 2 3 33 12 − 5 6 = ? 23 12 \dfrac{33}{12}-\dfrac{5}{6}\stackrel{?}{=}\dfrac{23}{12} 1 2 3 3 − 6 5 = ? 1 2 2 3 33 12 − 10 12 = ? 23 12 \dfrac{33}{12}-\dfrac{10}{12}\stackrel{?}{=}\dfrac{23}{12} 1 2 3 3 − 1 2 1 0 = ? 1 2 2 3
23 12 = 23 12 ✓ \dfrac{23}{12}=\dfrac{23}{12} \ {\color{#009600}{\huge{\checkmark}}} 1 2 2 3 = 1 2 2 3 ✓ 