Answer
$$\frac{x^2-5x}{x^2+3x}\frac{x+3}{x-5}=1$$
Explanation
| $$ \begin{aligned}\frac{x^2-5x}{x^2+3x}\frac{x+3}{x-5}& \xlongequal{ }\frac{x-5}{x+3}\frac{x+3}{x-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}1\end{aligned} $$ | |
| ① | Multiply $ \dfrac{x-5}{x+3} $ by $ \dfrac{x+3}{x-5} $ to get $ 1$. Step 1: Cancel $ \color{blue}{ x-5 } $ in first and second fraction. Step 2: Cancel $ \color{red}{ x+3 } $ in first and second fraction. Step 3: Multiply numerators and denominators. $$ \begin{aligned} \frac{x-5}{x+3} \cdot \frac{x+3}{x-5} & \xlongequal{\text{Step 1}} \frac{\color{blue}{1}}{x+3} \cdot \frac{x+3}{\color{blue}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{1}{\color{red}{1}} \cdot \frac{\color{red}{1}}{1} = \frac{1}{1} =1 \end{aligned} $$ |
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Simplify Rational Expressions