Solve 6-{4-[3\times-2+3\times6-(4\div-2+4-2)-9\div-3]+4}

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Question

$$6- \{ 4-[3 \times -2+3 \times 6-(4 \div -2+4-2)-9 \div -3]+4 \}$$

Evaluate

$13$

Solution Steps

Multiply $3$ and $-2$ to get $-6$. Multiply $3$ and $6$ to get $18$.

$$6-\left(4-\left(-6+18-\left(\frac{4}{-2}+4-2\right)-\frac{9}{-3}\right)+4\right)$$

Add $-6$ and $18$ to get $12$.

$$6-\left(4-\left(12-\left(\frac{4}{-2}+4-2\right)-\frac{9}{-3}\right)+4\right)$$

Divide $4$ by $-2$ to get $-2$.

$$6-\left(4-\left(12-\left(-2+4-2\right)-\frac{9}{-3}\right)+4\right)$$

Add $-2$ and $4$ to get $2$.

$$6-\left(4-\left(12-\left(2-2\right)-\frac{9}{-3}\right)+4\right)$$

Subtract $2$ from $2$ to get $0$.

$$6-\left(4-\left(12-0-\frac{9}{-3}\right)+4\right)$$

Subtract $0$ from $12$ to get $12$.

$$6-\left(4-\left(12-\frac{9}{-3}\right)+4\right)$$

Divide $9$ by $-3$ to get $-3$.

$$6-\left(4-\left(12-\left(-3\right)\right)+4\right)$$

The opposite of $-3$ is $3$.

$$6-\left(4-\left(12+3\right)+4\right)$$

Add $12$ and $3$ to get $15$.

$$6-\left(4-15+4\right)$$

Subtract $15$ from $4$ to get $-11$.

$$6-\left(-11+4\right)$$

Add $-11$ and $4$ to get $-7$.

$$6-\left(-7\right)$$

The opposite of $-7$ is $7$.

$$6+7$$

Add $6$ and $7$ to get $13$.

$$13$$

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