Answer
The solution set of the equation $\frac{x-1}{x-2}+\frac{x}{x-3}=\frac{1}{{{x}^{2}}-5x+6}$ is $\left\{ \frac{3\pm \sqrt{5}}{2} \right\}$.
Work Step by Step
The provided equation is $\frac{x-1}{x-2}+\frac{x}{x-3}=\frac{1}{{{x}^{2}}-5x+6}$.
Now factorize the expression ${{x}^{2}}-5x+6$ as,
$\begin{align}
& {{x}^{2}}-5x+6={{x}^{2}}-3x-2x+6 \\
& =x\left( x-3 \right)-2\left( x-3 \right) \\
& =\left( x-3 \right)\left( x-2 \right)
\end{align}$
The provided equation $\frac{x-1}{x-2}+\frac{x}{x-3}=\frac{1}{{{x}^{2}}-5x+6}$ can be written as,
$\frac{x-1}{x-2}+\frac{x}{x-3}=\frac{1}{\left( x-3 \right)\left( x-2 \right)}$
Multiply $\left( x-3 \right)\left( x-2 \right)$ on both sides.
$\left( x-3 \right)\left( x-2 \right)\left[ \frac{x-1}{x-2}+\frac{x}{x-3} \right]=\left( x-3 \right)\left( x-2 \right)\left[ \frac{1}{\left( x-3 \right)\left( x-2 \right)} \right]$
Use distributive property to simplify the above equation.
$\begin{align}
& \left( x-3 \right)\left( x-2 \right)\left[ \frac{x-1}{x-2}+\frac{x}{x-3} \right]=\left( x-3 \right)\left( x-2 \right)\left[ \frac{1}{\left( x-3 \right)\left( x-2 \right)} \right] \\
& \left( x-3 \right)\left( x-1 \right)+x\left( x-2 \right)=1 \\
& {{x}^{2}}-x-3x+3+{{x}^{2}}-2x=1 \\
& 2{{x}^{2}}-6x+3=1
\end{align}$
Subtract $1$ from both sides.
$2{{x}^{2}}-6x+2=0$
Use the formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve the $2{{x}^{2}}-6x+2=0$ equation.
Here, $a=2$, $b=-6$ and $c=2$.
The solution of the equation $2{{x}^{2}}-6x+2=0$ is,
$\begin{align}
& x=\frac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 2 \right)\left( 2 \right)}}{2\left( 2 \right)} \\
& =\frac{6\pm \sqrt{36-16}}{4} \\
& =\frac{6\pm \sqrt{20}}{4} \\
& =\frac{6\pm 2\sqrt{5}}{4}
\end{align}$
Further solve the above equation.
$\begin{align}
& x=\frac{6\pm 2\sqrt{5}}{4} \\
& =\frac{3\pm \sqrt{5}}{2}
\end{align}$
The solution set of the provided equation $\frac{x-1}{x-2}+\frac{x}{x-3}=\frac{1}{{{x}^{2}}-5x+6}$is $\left\{ \frac{3\pm \sqrt{5}}{2} \right\}$.
