Question
Answer
$$\dfrac{4}{2x+3}+\dfrac{x^2-x-2}{2x^2+5x+3}=\dfrac{x+2}{2x+3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4}{2x+3}+\frac{x^2-x-2}{2x^2+5x+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4}{2x+3}+\frac{x-2}{2x+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x+2}{2x+3}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^2-x-2}{2x^2+5x+3} $ to $ \dfrac{x-2}{2x+3} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+1}$. $$ \begin{aligned} \frac{x^2-x-2}{2x^2+5x+3} & =\frac{ \left( x-2 \right) \cdot \color{blue}{ \left( x+1 \right) }}{ \left( 2x+3 \right) \cdot \color{blue}{ \left( x+1 \right) }} = \\[1ex] &= \frac{x-2}{2x+3} \end{aligned} $$ |
| ② | Add $ \dfrac{4}{2x+3} $ and $ \dfrac{x-2}{2x+3} $ to get $ \dfrac{x+2}{2x+3} $. To add expressions with the same denominators, we add the numerators and write the result over the common denominator. $$ \begin{aligned} \frac{4}{2x+3} + \frac{x-2}{2x+3} & = \frac{4}{\color{blue}{2x+3}} + \frac{x-2}{\color{blue}{2x+3}} =\frac{ 4 + \left( x-2 \right) }{ \color{blue}{ 2x+3 }} = \\[1ex] &= \frac{x+2}{2x+3} \end{aligned} $$ |
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