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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Updated on: 10-Oct-2022

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