Question

Solve for $x:\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x},x\ne 0,1,2$

Answer 1

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

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

$=x(3x-5)=6(x-2)(x-1)$

$=3x^2-5x=6(x^2-x-2x+2)$

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

$=13x-3x^2-12=0$

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

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

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

$(3x-4)(x-3)$

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