Answer
$x = \frac{3±\sqrt{5}}{2}$
Work Step by Step
$\frac{x-1}{x-2} + \frac{x}{x-3} = \frac{1}{x^2-5x+6}$
${x^2-5x+6}$ can be rewritten as $(x-2)(x-3)$
Simplify by multiplying both sides by the least common multiple: $(x-2)(x-3)$
$\frac{x-1}{x-2} ⋅(x-2)(x-3)+ \frac{x}{x-3} ⋅(x-2)(x-3)= \frac{1}{(x-2)(x-3)}⋅(x-2)(x-3)$
$(x-1)(x-3) + x(x-2) = 1$
$x^2-4x+3+x^2-2x=1$
$2x^2-6x+3=1$
$2x^2-6x+2=0$
$2x^2-6x+2=0$
Solve by using the quadratic formula:
$x = \frac{-b±\sqrt{b^2-4ac}}{2a}$
$a=2$, $b=-6$, $c=2$
$x = \frac{-(-6)±\sqrt{(-6)^2-(4⋅2⋅2)}}{2(2)}$
$x = \frac{6±\sqrt{36-16}}{4}$
$x = \frac{6±\sqrt{20}}{4}$
$x = \frac{6±\sqrt{4⋅5}}{4}$
$x = \frac{6±2\sqrt{5}}{4}$
$x = \frac{3±\sqrt{5}}{2}$
