Question
Simplity:
$(\text{x}^2-\text{x})-\frac{1}{2}(\text{x}-3+3\text{x}^2)$
$(\text{x}^2-\text{x})-\frac{1}{2}(\text{x}-3+3\text{x}^2)$
✓
Answer
$(\text{x}^2-\text{x})-\frac{1}{2}(\text{x}-3+3\text{x}^2)$
$=\text{x}^2-\text{x}-\frac{1}{2}\text{x}+\frac{3}{2}-\frac{3}{2}\text{x}^2$
$=\text{x}^2-\frac{3}{2}\text{x}^2-\text{x}-\frac{1}{2}\text{x}+\frac{3}{2}$
$=\Big(1-\frac{3}{2}\Big)\text{x}^2-\Big(1+\frac{1}{2}\Big)\text{x}+\frac{3}{2}$
$=\frac{2-3}{2}\text{x}^2-\frac{2+1}{2}\text{x}+\frac{3}{2}$
$=-\frac{1}{2}\text{x}^2-\frac{3}{2}\text{x}+\frac{3}{2}$
$=\text{x}^2-\text{x}-\frac{1}{2}\text{x}+\frac{3}{2}-\frac{3}{2}\text{x}^2$
$=\text{x}^2-\frac{3}{2}\text{x}^2-\text{x}-\frac{1}{2}\text{x}+\frac{3}{2}$
$=\Big(1-\frac{3}{2}\Big)\text{x}^2-\Big(1+\frac{1}{2}\Big)\text{x}+\frac{3}{2}$
$=\frac{2-3}{2}\text{x}^2-\frac{2+1}{2}\text{x}+\frac{3}{2}$
$=-\frac{1}{2}\text{x}^2-\frac{3}{2}\text{x}+\frac{3}{2}$
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