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Question

Answer

$$\dfrac{2}{4}x+\dfrac{\dfrac{12}{4}}{2}x+6+\dfrac{1}{x}+3=\dfrac{2x^2+9x+1}{x}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}\frac{2}{4}x+\frac{\frac{12}{4}}{2}x+6+\frac{1}{x}+3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(\frac{2}{4}+\frac{\frac{12}{4}}{2})x+6+\frac{1}{x}+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1+3}{2}x+6+\frac{1}{x}+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4}{2}x+6+\frac{1}{x}+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{ 4 : \color{orangered}{ 2 } }{ 2 : \color{orangered}{ 2 }} \cdot x + 6 + \frac{1}{x} + 3 \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{2}{1}x+6+\frac{1}{x}+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}2x+6+\frac{1}{x}+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{2x^2+6x+1}{x}+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{2x^2+9x+1}{x}\end{aligned} $$
①
Use the distributive property.
②
$$ \frac{2}{4}+\frac{\frac{12}{4}}{2} = \frac{1}{2} \cdot \color{blue}{\frac{ 2 }{ 2}} + \frac{3}{2} \cdot \color{blue}{\frac{ 1 }{ 1}} = \frac{1+3}{2} $$
③

Simplify numerator

$$ \color{blue}{1} + \color{blue}{3} = \color{blue}{4} $$
④
Divide both the top and bottom numbers by $ \color{orangered}{ 2 } $.
⑤
Remove 1 from denominator.
⑥

Add $2x+6$ and $ \dfrac{1}{x} $ to get $ \dfrac{ \color{purple}{ 2x^2+6x+1 } }{ x }$.

Step 1: Write $ 2x+6 $ 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 $\color{blue}{ x }$.

$$ \begin{aligned} 2x+6+ \frac{1}{x} & \xlongequal{\text{Step 1}} \frac{2x+6}{\color{red}{1}} + \frac{1}{x} = \frac{ \left( 2x+6 \right) \cdot \color{blue}{ x }}{ 1 \cdot \color{blue}{ x }} + \frac{ 1 }{ x } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x^2+6x } }{ x } + \frac{ \color{purple}{ 1 } }{ x }=\frac{ \color{purple}{ 2x^2+6x+1 } }{ x } \end{aligned} $$
⑦

Add $ \dfrac{2x^2+6x+1}{x} $ and $ 3 $ to get $ \dfrac{ \color{purple}{ 2x^2+9x+1 } }{ x }$.

Step 1: Write $ 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 second fraction by $\color{blue}{x}$.

$$ \begin{aligned} \frac{2x^2+6x+1}{x} +3 & \xlongequal{\text{Step 1}} \frac{2x^2+6x+1}{x} + \frac{3}{\color{red}{1}} = \frac{ 2x^2+6x+1 }{ x } + \frac{ 3 \cdot \color{blue}{ x }}{ 1 \cdot \color{blue}{ x }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x^2+6x+1 } }{ x } + \frac{ \color{purple}{ 3x } }{ x }=\frac{ \color{purple}{ 2x^2+9x+1 } }{ x } \end{aligned} $$
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