Question

Solve the following quadratic equation by factorization.
$\frac{x+3}{x+2}=\frac{3x-7}{2x-3},x\ne -2,\frac{3}{2}$

Answer 1

$\frac{x+3}{x+2}=\frac{3x-7}{2x-3}$

Cross-multiplying we get;

$(x+3)(2x-3)=(3x-7)(x+2)$

$2x^2-3x+6x-9=3x^2+6x-7x-14$

$2x^2+3x-9=3x^2-x-14$

$x^2-4x-5=0$

$x^2+x-5x-5=0$

$x(x+1)-5(x+1)=0$

$(x+1)(x-5)=0$

Therefore,

$x=-1,5$