Solve the following:$\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}$

Given: 

The given expression is $\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}$.

To do :

We have to find the value of x.

Solution :

$\frac{1}{x-2}+\frac{2}{x-1} =\frac{6}{x}$

$\frac{(x-1)+2(x-2)}{(x-2)(x-1)}=\frac{6}{x}$

$\frac{(x-1+2 x-4)}{(x^{2}-x-2 x+2)}=\frac{6}{x}$

$\frac{3 x-5}{x^{2}-3 x+2}=\frac{6}{x}$

$3 x^{2}-5 x=6 x^{2}-18 x+12$

$3 x^{2}-6 x^{2}=5 x-18 x+12$

$-3 x^{2}=-13 x+12$

$3 x^{2}-13 x+12=0$

$3 x^{2}-9 x-4 x+12=0$

$3 x(x-3)-4(x-3)=0$

$(3 x-4)(x-3)=0$

$(3 x-4)=0$, $x=\frac{4}{3}$

$(x-3)=0$, $x=3$

Therefore,The value of x is $\frac{4}{3}$ or 3.

Tutorialspoint

e

Updated on: 10-Oct-2022

36 Views

Kickstart Your Career

Get certified by completing the course

Get Started