Question

Solve by factorization method:

$\left(\frac{4}{x}\right)-3=\frac{5}{\left(2x+3\right)},x\ne 0,\frac{-3}{2}$

Answer 1

Given expression is,

$\left(\frac{4}{x}\right)-3=\frac{5}{\left(2x+3\right)},x\ne 0,\frac{-3}{2}$

On simplifying the given expression, we get;

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

$\Rightarrow$ $\left(4-3x\right)\left(2x+3\right)=5x$

$\Rightarrow$ $8x-12-6x^2-9x=5x$

$$\begin{gathered} \Rightarrow 6x^2+9x-8x+12+5x=0 \\ \Rightarrow 6x^2-6x+12=0 \end{gathered}$$

$\Rightarrow$ $x^2+x-2=0$ [Divide the equation by $6$]

By following factorization method,

$$\begin{gathered} x^2+2x-x-2=0 \\ \Rightarrow x\left(x+2\right)-1\left(x+2\right)=0 \\ \Rightarrow \left(x+2\right)\left(x-1\right)=0 \\ \Rightarrow x+2=0orx-1=0 \\ \Rightarrow x=-2,1 \end{gathered}$$

Hence, the solution of the given expression is $x=-2,x=1$