Question
Answer
$$2\cdot\dfrac{x-1}{x-1}(x+3)=2x+6$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2 \cdot \frac{x-1}{x-1}(x+3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2\cdot1(x+3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2x+6\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x-1}{x-1} $ to $ 1$. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-1}$. $$ \begin{aligned} \frac{x-1}{x-1} & =\frac{ 1 \cdot \color{blue}{ \left( x-1 \right) }}{ 1 \cdot \color{blue}{ \left( x-1 \right) }} = \\[1ex] &= \frac{1}{1} =1 \end{aligned} $$ |
| ② | Multiply $ \color{blue}{2} $ by $ \left( x+3\right) $ $$ \color{blue}{2} \cdot \left( x+3\right) = 2x+6 $$ |
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