## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

The simplified form of the expression $\frac{1}{5-x}+\frac{x}{2x-10}$ is $\frac{\left( x-2 \right)}{2\left( x-5 \right)}$.
$\frac{1}{5-x}+\frac{x}{2x-10}$ Obtain the alternative form of the expression by multiplying the second term of rational expression $\frac{1}{5-x}+\frac{x}{2x-10}$ by 1 in the form$\frac{-1}{-1}$: \begin{align} & \frac{1}{5-x}+\frac{x}{2x-10}=\frac{1}{5-x}+\frac{x}{2\left( x-5 \right)}.1 \\ & \frac{1}{5-x}+\frac{x}{2\left( x-5 \right)}.\frac{-1}{-1} \end{align} Apply the Distributive property: \begin{align} & \frac{1}{5-x}+\frac{x}{2x-10}=\frac{1}{5-x}+\frac{x\left( -1 \right)}{2\left( x-5 \right)\left( -1 \right)} \\ & =\frac{1}{5-x}+\frac{-\left( x \right)}{-2\left( x-5 \right)} \\ & =\frac{1}{5-x}-\frac{x}{2\left( 5-x \right)} \end{align} \begin{align} & \frac{1}{5-x}=\frac{1\cdot 2}{2\left( 5-x \right)} \\ & =\frac{2}{2\left( 5-x \right)} \end{align} And $\frac{x}{2\left( x-5 \right)}=\frac{x}{2\left( x-5 \right)}$ Thus, expression becomes: $\frac{1}{5-x}+\frac{x}{2x-10}=\frac{2}{2\left( 5-x \right)}-\frac{x}{2\left( x-5 \right)}$ Now, the denominators are same. So, subtract the numerators and keep the common denominator: $\frac{1}{5-x}+\frac{x}{2x-10}=\frac{2-x}{2\left( 5-x \right)}$ Rearrange the terms and simplify further as follows: \begin{align} & \frac{1}{5-x}+\frac{x}{2x-10}=\frac{2-x}{2\left( 5-x \right)} \\ & =\frac{x-2}{2\left( x-5 \right)} \end{align}