Answer
$$x^2+\frac{8}{4}x^2-4x-3=3x^2-4x-3$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}x^2+\frac{8}{4}x^2-4x-3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^2 + \frac{ 8 : \color{orangered}{ 4 } }{ 4 : \color{orangered}{ 4 }} \cdot x^2 - 4x - 3 \xlongequal{ } \\[1 em] & \xlongequal{ }x^2+\frac{2}{1}x^2-4x-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^2+2x^2-4x-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}3x^2-4x-3\end{aligned} $$ | |
| ① | Divide both the top and bottom numbers by $ \color{orangered}{ 4 } $. |
| ② | Remove 1 from denominator. |
| ③ | Combine like terms: $$ \color{blue}{x^2} + \color{blue}{2x^2} = \color{blue}{3x^2} $$ |
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