## Precalculus (6th Edition) Blitzer

The solution set of the equation $\frac{x-1}{x-2}+\frac{x}{x-3}=\frac{1}{{{x}^{2}}-5x+6}$ is $\left\{ \frac{3\pm \sqrt{5}}{2} \right\}$.
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 $\left\{ \frac{3\pm \sqrt{5}}{2} \right\}$.