Answer:
(C)x=3
Step-by-step explanation:
We want to solve for x in: [tex]\dfrac{x}{x-1} -\dfrac{1}{x-2}=\dfrac{2x-5}{x^2-3x+2}[/tex]
Take the LCM of the Left Hand Side
[tex]\dfrac{x(x-2)-1(x-1)}{(x-1)(x-2)} =\dfrac{2x-5}{x^2-3x+2}\\\dfrac{x(x-2)-1(x-1)}{x^2-3x+2} =\dfrac{2x-5}{x^2-3x+2}\\$Since we have the same denominator, then:\\x(x-2)-1(x-1)=2x-5\\x^2-2x-x+1-2x+5=0\\x^2-5x+6=0\\$Factorize\\x^2-3x-2x+6=0\\x(x-3)-2(x-3)=0\\(x-3)(x-2)=0\\x-3=0$ or $x-2=0\\x=3$ or x=2[/tex]
However, substituting x=2 does not satisfy the given equation as it would make it undefined.
Therefore x=3
