Question

Solve each of the following quadratic equations:
$\frac{4}{x}-3=\frac{5}{2x+3},x\ne 0,\frac{-3}{2}$

Answer 1

$\frac{4}{x}-3=\frac{5}{2x+3}$

⇒$\frac{(4-3x)}{x}=\frac{5}{2x+3}$

⇒$(4-3x)(2x+3)=5x$

⇒$8x+12-6x^2-9x=5x$

⇒$6x^2+6x-12=0$

⇒$x^2+x-2=0$

⇒$x^2+(2-1)x-2=0$

⇒$x^2+2x-x-2=0$

⇒$x(x+2)-1(x+2)=0$

⇒$(x+2)(x-1)=0$

when

$x+2=0$

$x=-2$

and

$x-1=0$

$x=1$

So x= 1 and -2