Math Calculators, Lessons and Formulas

It is time to solve your math problem

mathportal.org

◀ back to index

Question

Answer

$$3y^3+\dfrac{81}{y^2}-3y+9=\dfrac{3y^5-3y^3+9y^2+81}{y^2}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}3y^3+\frac{81}{y^2}-3y+9& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3y^5+81}{y^2}-3y+9 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3y^5-3y^3+81}{y^2}+9 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3y^5-3y^3+9y^2+81}{y^2}\end{aligned} $$
①

Add $3y^3$ and $ \dfrac{81}{y^2} $ to get $ \dfrac{ \color{purple}{ 3y^5+81 } }{ y^2 }$.

Step 1: Write $ 3y^3 $ 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 first fraction by $\color{blue}{ y^2 }$.

$$ \begin{aligned} 3y^3+ \frac{81}{y^2} & \xlongequal{\text{Step 1}} \frac{3y^3}{\color{red}{1}} + \frac{81}{y^2} = \frac{ 3y^3 \cdot \color{blue}{ y^2 }}{ 1 \cdot \color{blue}{ y^2 }} + \frac{ 81 }{ y^2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 3y^5 } }{ y^2 } + \frac{ \color{purple}{ 81 } }{ y^2 }=\frac{ \color{purple}{ 3y^5+81 } }{ y^2 } \end{aligned} $$
②

Subtract $3y$ from $ \dfrac{3y^5+81}{y^2} $ to get $ \dfrac{ \color{purple}{ 3y^5-3y^3+81 } }{ y^2 }$.

Step 1: Write $ 3y $ 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 $\color{blue}{y^2}$.

$$ \begin{aligned} \frac{3y^5+81}{y^2} -3y & \xlongequal{\text{Step 1}} \frac{3y^5+81}{y^2} - \frac{3y}{\color{red}{1}} = \frac{ 3y^5+81 }{ y^2 } - \frac{ 3y \cdot \color{blue}{ y^2 }}{ 1 \cdot \color{blue}{ y^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 3y^5+81 } }{ y^2 } - \frac{ \color{purple}{ 3y^3 } }{ y^2 }=\frac{ \color{purple}{ 3y^5-3y^3+81 } }{ y^2 } \end{aligned} $$
③

Add $ \dfrac{3y^5-3y^3+81}{y^2} $ and $ 9 $ to get $ \dfrac{ \color{purple}{ 3y^5-3y^3+9y^2+81 } }{ y^2 }$.

Step 1: Write $ 9 $ 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 $\color{blue}{y^2}$.

$$ \begin{aligned} \frac{3y^5-3y^3+81}{y^2} +9 & \xlongequal{\text{Step 1}} \frac{3y^5-3y^3+81}{y^2} + \frac{9}{\color{red}{1}} = \frac{ 3y^5-3y^3+81 }{ y^2 } + \frac{ 9 \cdot \color{blue}{ y^2 }}{ 1 \cdot \color{blue}{ y^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 3y^5-3y^3+81 } }{ y^2 } + \frac{ \color{purple}{ 9y^2 } }{ y^2 }=\frac{ \color{purple}{ 3y^5-3y^3+9y^2+81 } }{ y^2 } \end{aligned} $$
This page was created using
Simplify Rational Expressions