Question

How to solve this linear equation step by step please

Answer 1

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

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