Question

Solve the following equations and verify your answer :

$\frac{(x+2)(2x-3)-2x^2+6}{x-5}=2$

Answer 1

$\frac{(x+2)(2x-3)-2x^2+6}{x-5}=\frac{2}{1}$
By cross multiplication
$(x+2)(2x-3)-2x^2+6=2(x-5)2x^2-3x+4x-6-2x^2+6=2x-10x=2x-10\Rightarrow x-2x=-10\Rightarrow -x=-10\Rightarrow x=10\therefore x=10$
Verification,
$L.H.S.=\frac{(x+2)(2x-3)-2x^2+6}{x-5}=\frac{(10+2)(2\times 10-3)-2(10{)}^2+6}{10-5}=\frac{12(20-3)-2\times \left(100\right)+6}{5}=\frac{12\times 17-200+6}{5}=\frac{204-200+6}{5}=\frac{10}{5}=2=R.H.S.$