Solve the following quadratic equation by factorization:
$\frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1}, x ≠-1, \frac{1}{3}$
Given:
Given quadratic equation is $\frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1}, x ≠-1, \frac{1}{3}$.
To do:
We have to solve the given quadratic equation by factorization.
Solution:
$\frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1}$
$\frac{3(2)-1(x+1)}{(x+1)(2)}=\frac{2}{3x-1}$
$\frac{6-x-1}{2x+2}=\frac{2}{3x-1}$
$\frac{5-x}{2x+2}=\frac{2}{3x-1}$
$(5-x)(3x-1)=2(2x+2)$ (on cross multiplication)
$15x-5-3x^2+x=4x+4$
$-3x^2+16x-5=4x+4$
$3x^2+4x-16x+4+5=0$
$3x^2-12x+9=0$
$3x^2-3x-9x+9=0$
$3x(x-1)-9(x-1)=0$
$(3x-9)(x-1)=0$
$3x-9=0$ or $x-1=0$
$3x-9=0$ or $x-1=0$
$3x=9$ or $x=1$
$x=\frac{9}{3}$ or $x=1$
$x=3$ or $x=1$
The values of $x$ are $1$ and $3$.  
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