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Question

Answer

$$\dfrac{y}{y}+8-8y-9=-\dfrac{8y^2}{y}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}\frac{y}{y}+8-8y-9& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{9y}{y}-8y-9 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-8y^2+9y}{y}-9 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-\frac{8y^2}{y}\end{aligned} $$
①

Add $ \dfrac{y}{y} $ and $ 8 $ to get $ \dfrac{ \color{purple}{ 9y } }{ y }$.

Step 1: Write $ 8 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{y}$.

$$ \begin{aligned} \frac{y}{y} +8 & \xlongequal{\text{Step 1}} \frac{y}{y} + \frac{8}{\color{red}{1}} = \frac{ y }{ y } + \frac{ 8 \cdot \color{blue}{ y }}{ 1 \cdot \color{blue}{ y }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ y } }{ y } + \frac{ \color{purple}{ 8y } }{ y }=\frac{ \color{purple}{ 9y } }{ y } \end{aligned} $$
②

Subtract $8y$ from $ \dfrac{9y}{y} $ to get $ \dfrac{ \color{purple}{ -8y^2+9y } }{ y }$.

Step 1: Write $ 8y $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{y}$.

$$ \begin{aligned} \frac{9y}{y} -8y & \xlongequal{\text{Step 1}} \frac{9y}{y} - \frac{8y}{\color{red}{1}} = \frac{ 9y }{ y } - \frac{ 8y \cdot \color{blue}{ y }}{ 1 \cdot \color{blue}{ y }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 9y } }{ y } - \frac{ \color{purple}{ 8y^2 } }{ y }=\frac{ \color{purple}{ -8y^2+9y } }{ y } \end{aligned} $$
③

Subtract $9$ from $ \dfrac{-8y^2+9y}{y} $ to get $ \dfrac{ \color{purple}{ -8y^2 } }{ y }$.

Step 1: Write $ 9 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{y}$.

$$ \begin{aligned} \frac{-8y^2+9y}{y} -9 & \xlongequal{\text{Step 1}} \frac{-8y^2+9y}{y} - \frac{9}{\color{red}{1}} = \frac{ -8y^2+9y }{ y } - \frac{ 9 \cdot \color{blue}{ y }}{ 1 \cdot \color{blue}{ y }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -8y^2+9y } }{ y } - \frac{ \color{purple}{ 9y } }{ y }=\frac{ \color{purple}{ -8y^2 } }{ y } \end{aligned} $$
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