Tap the blue circles to see an explanation.
$$ \begin{aligned}\frac{3(d+8)-3}{7(d+8)-7}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3d+24-3}{7d+56-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3d+21}{7d+56-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{3d+21}{7d+49} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{3}{7}\end{aligned} $$ | |
① | Multiply $ \color{blue}{3} $ by $ \left( d+8\right) $ $$ \color{blue}{3} \cdot \left( d+8\right) = 3d+24 $$ |
② | Multiply $ \color{blue}{7} $ by $ \left( d+8\right) $ $$ \color{blue}{7} \cdot \left( d+8\right) = 7d+56 $$ |
③ | $$ 3d+ \color{blue}{24} \color{blue}{-3} = 3d+ \color{blue}{21} $$ |
④ | $$ 7d+ \color{blue}{56} \color{blue}{-7} = 7d+ \color{blue}{49} $$ |
⑤ | Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{d+7}$. $$ \begin{aligned} \frac{3d+21}{7d+49} & =\frac{ 3 \cdot \color{blue}{ \left( d+7 \right) }}{ 7 \cdot \color{blue}{ \left( d+7 \right) }} = \\[1ex] &= \frac{3}{7} \end{aligned} $$ |