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Question

Answer

$$x^2-8x+\dfrac{15}{10}+8x-2x^2=\dfrac{-2x^2+3}{2}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}x^2-8x+\frac{15}{10}+8x-2x^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^2-8x + \frac{ 15 : \color{orangered}{ 5 } }{ 10 : \color{orangered}{ 5 }} + 8x - 2x^2 \xlongequal{ } \\[1 em] & \xlongequal{ }x^2-8x+\frac{3}{2}+8x-2x^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2x^2-16x+3}{2}+8x-2x^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2x^2+3}{2}-2x^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-2x^2+3}{2}\end{aligned} $$
①
Divide both the top and bottom numbers by $ \color{orangered}{ 5 } $.
②

Add $x^2-8x$ and $ \dfrac{3}{2} $ to get $ \dfrac{ \color{purple}{ 2x^2-16x+3 } }{ 2 }$.

Step 1: Write $ x^2-8x $ 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 $\color{blue}{ 2 }$.

$$ \begin{aligned} x^2-8x+ \frac{3}{2} & \xlongequal{\text{Step 1}} \frac{x^2-8x}{\color{red}{1}} + \frac{3}{2} = \frac{ \left( x^2-8x \right) \cdot \color{blue}{ 2 }}{ 1 \cdot \color{blue}{ 2 }} + \frac{ 3 }{ 2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x^2-16x } }{ 2 } + \frac{ \color{purple}{ 3 } }{ 2 }=\frac{ \color{purple}{ 2x^2-16x+3 } }{ 2 } \end{aligned} $$
③

Add $ \dfrac{2x^2-16x+3}{2} $ and $ 8x $ to get $ \dfrac{ \color{purple}{ 2x^2+3 } }{ 2 }$.

Step 1: Write $ 8x $ 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 $\color{blue}{2}$.

$$ \begin{aligned} \frac{2x^2-16x+3}{2} +8x & \xlongequal{\text{Step 1}} \frac{2x^2-16x+3}{2} + \frac{8x}{\color{red}{1}} = \frac{ 2x^2-16x+3 }{ 2 } + \frac{ 8x \cdot \color{blue}{ 2 }}{ 1 \cdot \color{blue}{ 2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x^2-16x+3 } }{ 2 } + \frac{ \color{purple}{ 16x } }{ 2 }=\frac{ \color{purple}{ 2x^2+3 } }{ 2 } \end{aligned} $$
④

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

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

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