Question
Answer
$$7x^2y^3-2-(3x^3y^2+x^2y^2-6)=-3x^3y^2+7x^2y^3-x^2y^2+4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}7x^2y^3-2-(3x^3y^2+x^2y^2-6)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}7x^2y^3-2-3x^3y^2-x^2y^2+6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-3x^3y^2+7x^2y^3-x^2y^2+4\end{aligned} $$ | |
| ① | Remove the parentheses by changing the sign of each term within them. $$ - \left( 3x^3y^2+x^2y^2-6 \right) = -3x^3y^2-x^2y^2+6 $$ |
| ② | Combine like terms: $$ 7x^2y^3 \color{blue}{-2} -3x^3y^2-x^2y^2+ \color{blue}{6} = -3x^3y^2+7x^2y^3-x^2y^2+ \color{blue}{4} $$ |
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