Math Calculators, Lessons and Formulas

It is time to solve your math problem

mathportal.org

back to index

Answer

$$2xj^4-3yj^3+\frac{(1-2j)^2}{j^4}+j^5=\frac{j^9+2j^8x-3j^7y+4j^2-4j+1}{j^4}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}2xj^4-3yj^3+\frac{(1-2j)^2}{j^4}+j^5& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2xj^4-3yj^3+\frac{1-4j+4j^2}{j^4}+j^5 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2j^8x-3j^7y+4j^2-4j+1}{j^4}+j^5 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{j^9+2j^8x-3j^7y+4j^2-4j+1}{j^4}\end{aligned} $$
①

Find $ \left(1-2j\right)^2 $ using formula.

$$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ 1 } $ and $ B = \color{red}{ 2j }$.

$$ \begin{aligned}\left(1-2j\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot 2j + \color{red}{\left( 2j \right)^2} = 1-4j+4j^2\end{aligned} $$
②

Add $2j^4x-3j^3y$ and $ \dfrac{1-4j+4j^2}{j^4} $ to get $ \dfrac{ \color{purple}{ 2j^8x-3j^7y+4j^2-4j+1 } }{ j^4 }$.

Step 1: Write $ 2j^4x-3j^3y $ 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 $\color{blue}{ j^4 }$.

$$ \begin{aligned} 2j^4x-3j^3y+ \frac{1-4j+4j^2}{j^4} & \xlongequal{\text{Step 1}} \frac{2j^4x-3j^3y}{\color{red}{1}} + \frac{1-4j+4j^2}{j^4} = \frac{ \left( 2j^4x-3j^3y \right) \cdot \color{blue}{ j^4 }}{ 1 \cdot \color{blue}{ j^4 }} + \frac{ 1-4j+4j^2 }{ j^4 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2j^8x-3j^7y } }{ j^4 } + \frac{ \color{purple}{ 1-4j+4j^2 } }{ j^4 }=\frac{ \color{purple}{ 2j^8x-3j^7y+4j^2-4j+1 } }{ j^4 } \end{aligned} $$
③

Add $ \dfrac{2j^8x-3j^7y+4j^2-4j+1}{j^4} $ and $ j^5 $ to get $ \dfrac{ \color{purple}{ j^9+2j^8x-3j^7y+4j^2-4j+1 } }{ j^4 }$.

Step 1: Write $ j^5 $ 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 $\color{blue}{j^4}$.

$$ \begin{aligned} \frac{2j^8x-3j^7y+4j^2-4j+1}{j^4} +j^5 & \xlongequal{\text{Step 1}} \frac{2j^8x-3j^7y+4j^2-4j+1}{j^4} + \frac{j^5}{\color{red}{1}} = \\[1ex] &= \frac{ 2j^8x-3j^7y+4j^2-4j+1 }{ j^4 } + \frac{ j^5 \cdot \color{blue}{ j^4 }}{ 1 \cdot \color{blue}{ j^4 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2j^8x-3j^7y+4j^2-4j+1 } }{ j^4 } + \frac{ \color{purple}{ j^9 } }{ j^4 }=\frac{ \color{purple}{ j^9+2j^8x-3j^7y+4j^2-4j+1 } }{ j^4 } \end{aligned} $$
This page was created using
Complex Expression calculator