How to solve this linear equation step by step please

Mathematics

1 answer:

mote1985 [20]2 years ago

7 0

$\bf \cfrac{3}{c^2-4c}-\cfrac{9}{2c^2+3c}=\cfrac{2}{2c^2-5c-12}\\\\\\ 
\begin{cases}
c^2-4c=&c(c-4)\\
2c^2+3c=&c(2c+3)\\\\
2c^2-5c-12=
&(2c+3)(c-4)
\end{cases}\\\\
-------------------------------\\\\
\textit{therefore, our LCD for the \underline{left-side} will have to be }\underline{c(c-4)(2c+3)}\\\\
-------------------------------\\\\$

$\bf \cfrac{3}{c(c-4)}-\cfrac{9}{c(2c+3)}=\cfrac{2}{(2c+3)(c-4)}
\\\\\\
\cfrac{3(2c+3)-9(c-4)}{c(c-4)(2c+3)}=\cfrac{2}{(2c+3)(c-4)}
\\\\\\
\cfrac{6c+9-(9c-36)}{c(c-4)(2c+3)}=\cfrac{2}{(2c+3)(c-4)}
\\\\\\
\cfrac{6c+9-9c+36}{1}=\cfrac{2[c(c-4)(2c+3)]}{(2c+3)(c-4)}
\\\\\\
-3c+45=2c\implies 45=5c\implies \cfrac{45}{5}=c\implies 9=c$

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