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Question

Answer

$$\dfrac{x^4}{7}+3x-7x^2-4=\dfrac{x^4-49x^2+21x-28}{7}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}\frac{x^4}{7}+3x-7x^2-4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^4+21x}{7}-7x^2-4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^4-49x^2+21x}{7}-4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{x^4-49x^2+21x-28}{7}\end{aligned} $$
①

Add $ \dfrac{x^4}{7} $ and $ 3x $ to get $ \dfrac{ \color{purple}{ x^4+21x } }{ 7 }$.

Step 1: Write $ 3x $ 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 $\color{blue}{7}$.

$$ \begin{aligned} \frac{x^4}{7} +3x & \xlongequal{\text{Step 1}} \frac{x^4}{7} + \frac{3x}{\color{red}{1}} = \frac{ x^4 }{ 7 } + \frac{ 3x \cdot \color{blue}{ 7 }}{ 1 \cdot \color{blue}{ 7 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^4 } }{ 7 } + \frac{ \color{purple}{ 21x } }{ 7 }=\frac{ \color{purple}{ x^4+21x } }{ 7 } \end{aligned} $$
②

Subtract $7x^2$ from $ \dfrac{x^4+21x}{7} $ to get $ \dfrac{ \color{purple}{ x^4-49x^2+21x } }{ 7 }$.

Step 1: Write $ 7x^2 $ 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 $\color{blue}{7}$.

$$ \begin{aligned} \frac{x^4+21x}{7} -7x^2 & \xlongequal{\text{Step 1}} \frac{x^4+21x}{7} - \frac{7x^2}{\color{red}{1}} = \frac{ x^4+21x }{ 7 } - \frac{ 7x^2 \cdot \color{blue}{ 7 }}{ 1 \cdot \color{blue}{ 7 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^4+21x } }{ 7 } - \frac{ \color{purple}{ 49x^2 } }{ 7 }=\frac{ \color{purple}{ x^4-49x^2+21x } }{ 7 } \end{aligned} $$
③

Subtract $4$ from $ \dfrac{x^4-49x^2+21x}{7} $ to get $ \dfrac{ \color{purple}{ x^4-49x^2+21x-28 } }{ 7 }$.

Step 1: Write $ 4 $ 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 $\color{blue}{7}$.

$$ \begin{aligned} \frac{x^4-49x^2+21x}{7} -4 & \xlongequal{\text{Step 1}} \frac{x^4-49x^2+21x}{7} - \frac{4}{\color{red}{1}} = \frac{ x^4-49x^2+21x }{ 7 } - \frac{ 4 \cdot \color{blue}{ 7 }}{ 1 \cdot \color{blue}{ 7 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^4-49x^2+21x } }{ 7 } - \frac{ \color{purple}{ 28 } }{ 7 }=\frac{ \color{purple}{ x^4-49x^2+21x-28 } }{ 7 } \end{aligned} $$
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