Answer
The solution is $x=\dfrac{2}{3}$
Work Step by Step
$\dfrac{1}{x-2}-\dfrac{3x}{x-1}=\dfrac{2x+1}{x^{2}-3x+2}$
Factor the denominator of the fraction on the right side of the equation:
$\dfrac{1}{x-2}-\dfrac{3x}{x-1}=\dfrac{2x+1}{(x-2)(x-1)}$
Multiply the whole equation by $(x-2)(x-1)$:
$(x-2)(x-1)\Big[\dfrac{1}{x-2}-\dfrac{3x}{x-1}=\dfrac{2x+1}{(x-2)(x-1)}\Big]$
$x-1-3x(x-2)=2x+1$
$x-1-3x^{2}+6x=2x+1$
Take all terms to the right side and simplify:
$0=3x^{2}+2x-6x-x+1+1$
$0=3x^{2}-5x+2$
Rearrange:
$3x^{2}-5x+2=0$
Solve by factoring:
$(3x-2)(x-1)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$3x-2=0$
$3x=2$
$x=\dfrac{2}{3}$
$x-1=0$
$x=1$
The original equation is undefined for $x=1$, so the solution is $x=\dfrac{2}{3}$
