Answer
$$\frac{5x^2-5}{(x-1)(x+5)}=\frac{5x+5}{x+5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5x^2-5}{(x-1)(x+5)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5x^2-5}{x^2+5x-x-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{5x^2-5}{x^2+4x-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{5x+5}{x+5}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x-1}\right) $ by each term in $ \left( x+5\right) $. $$ \left( \color{blue}{x-1}\right) \cdot \left( x+5\right) = x^2+5x-x-5 $$ |
| ② | Simplify denominator $$ x^2+ \color{blue}{5x} \color{blue}{-x} -5 = x^2+ \color{blue}{4x} -5 $$ |
| ③ | Simplify $ \dfrac{5x^2-5}{x^2+4x-5} $ to $ \dfrac{5x+5}{x+5} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-1}$. $$ \begin{aligned} \frac{5x^2-5}{x^2+4x-5} & =\frac{ \left( 5x+5 \right) \cdot \color{blue}{ \left( x-1 \right) }}{ \left( x+5 \right) \cdot \color{blue}{ \left( x-1 \right) }} = \\[1ex] &= \frac{5x+5}{x+5} \end{aligned} $$ |
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