Question

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

Find the value of 'x'

• A. $x=3or,x=\frac{4}{3}$
  
• B. $x=3or,x=\frac{4}{7}$
  
• C. $x=4or,x=\frac{4}{3}$
  
• D. $x=7or,x=\frac{4}{3}$
  

Answer 1

The correct option is A

$x=3or,x=\frac{4}{3}$

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

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

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

$\Rightarrow 3x^2-5x=6x^2-18x+12$

$\Rightarrow 3x^2-13x+12=0$

$\Rightarrow 3x^2-9x-4x+12=0$

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

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

$\Rightarrow x-3=0or,3x-4=0$

$\Rightarrow x=3or,x=\frac{4}{3}$