Question
Answer
$$\dfrac{x^2+x-30}{(x-7)(x+3)}=\dfrac{x^2+x-30}{x^2-4x-21}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{x^2+x-30}{(x-7)(x+3)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^2+x-30}{x^2+3x-7x-21} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^2+x-30}{x^2-4x-21}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x-7}\right) $ by each term in $ \left( x+3\right) $. $$ \left( \color{blue}{x-7}\right) \cdot \left( x+3\right) = x^2+3x-7x-21 $$ |
| ② | Simplify denominator $$ x^2+ \color{blue}{3x} \color{blue}{-7x} -21 = x^2 \color{blue}{-4x} -21 $$ |
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Simplify Rational Expressions