Solve -8-{5-[6-(4-3\div-6+5-2)]+7}-6\div-3

Reduce the fraction $\frac{3}{-6}$ to lowest terms by extracting and canceling out $3$.

$$-8-\left(5-\left(6-\left(4-\left(-\frac{1}{2}\right)+5-2\right)\right)+7\right)-\frac{6}{-3}$$

The opposite of $-\frac{1}{2}$ is $\frac{1}{2}$.

$$-8-\left(5-\left(6-\left(4+\frac{1}{2}+5-2\right)\right)+7\right)-\frac{6}{-3}$$

Convert $4$ to fraction $\frac{8}{2}$.

$$-8-\left(5-\left(6-\left(\frac{8}{2}+\frac{1}{2}+5-2\right)\right)+7\right)-\frac{6}{-3}$$

Since $\frac{8}{2}$ and $\frac{1}{2}$ have the same denominator, add them by adding their numerators.

$$-8-\left(5-\left(6-\left(\frac{8+1}{2}+5-2\right)\right)+7\right)-\frac{6}{-3}$$

Add $8$ and $1$ to get $9$.

$$-8-\left(5-\left(6-\left(\frac{9}{2}+5-2\right)\right)+7\right)-\frac{6}{-3}$$

Convert $5$ to fraction $\frac{10}{2}$.

$$-8-\left(5-\left(6-\left(\frac{9}{2}+\frac{10}{2}-2\right)\right)+7\right)-\frac{6}{-3}$$

Since $\frac{9}{2}$ and $\frac{10}{2}$ have the same denominator, add them by adding their numerators.

$$-8-\left(5-\left(6-\left(\frac{9+10}{2}-2\right)\right)+7\right)-\frac{6}{-3}$$

Add $9$ and $10$ to get $19$.

$$-8-\left(5-\left(6-\left(\frac{19}{2}-2\right)\right)+7\right)-\frac{6}{-3}$$

Convert $2$ to fraction $\frac{4}{2}$.

$$-8-\left(5-\left(6-\left(\frac{19}{2}-\frac{4}{2}\right)\right)+7\right)-\frac{6}{-3}$$

Since $\frac{19}{2}$ and $\frac{4}{2}$ have the same denominator, subtract them by subtracting their numerators.

$$-8-\left(5-\left(6-\frac{19-4}{2}\right)+7\right)-\frac{6}{-3}$$

Subtract $4$ from $19$ to get $15$.

$$-8-\left(5-\left(6-\frac{15}{2}\right)+7\right)-\frac{6}{-3}$$

Convert $6$ to fraction $\frac{12}{2}$.

$$-8-\left(5-\left(\frac{12}{2}-\frac{15}{2}\right)+7\right)-\frac{6}{-3}$$

Since $\frac{12}{2}$ and $\frac{15}{2}$ have the same denominator, subtract them by subtracting their numerators.

$$-8-\left(5-\frac{12-15}{2}+7\right)-\frac{6}{-3}$$

Subtract $15$ from $12$ to get $-3$.

$$-8-\left(5-\left(-\frac{3}{2}\right)+7\right)-\frac{6}{-3}$$

The opposite of $-\frac{3}{2}$ is $\frac{3}{2}$.

$$-8-\left(5+\frac{3}{2}+7\right)-\frac{6}{-3}$$

Convert $5$ to fraction $\frac{10}{2}$.

$$-8-\left(\frac{10}{2}+\frac{3}{2}+7\right)-\frac{6}{-3}$$

Since $\frac{10}{2}$ and $\frac{3}{2}$ have the same denominator, add them by adding their numerators.

$$-8-\left(\frac{10+3}{2}+7\right)-\frac{6}{-3}$$

Add $10$ and $3$ to get $13$.

$$-8-\left(\frac{13}{2}+7\right)-\frac{6}{-3}$$

Convert $7$ to fraction $\frac{14}{2}$.

$$-8-\left(\frac{13}{2}+\frac{14}{2}\right)-\frac{6}{-3}$$

Since $\frac{13}{2}$ and $\frac{14}{2}$ have the same denominator, add them by adding their numerators.

$$-8-\frac{13+14}{2}-\frac{6}{-3}$$

Add $13$ and $14$ to get $27$.

$$-8-\frac{27}{2}-\frac{6}{-3}$$

Convert $-8$ to fraction $-\frac{16}{2}$.

$$-\frac{16}{2}-\frac{27}{2}-\frac{6}{-3}$$

Since $-\frac{16}{2}$ and $\frac{27}{2}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{-16-27}{2}-\frac{6}{-3}$$

Subtract $27$ from $-16$ to get $-43$.

$$-\frac{43}{2}-\frac{6}{-3}$$

Divide $6$ by $-3$ to get $-2$.

$$-\frac{43}{2}-\left(-2\right)$$

The opposite of $-2$ is $2$.

$$-\frac{43}{2}+2$$

Convert $2$ to fraction $\frac{4}{2}$.

$$-\frac{43}{2}+\frac{4}{2}$$

Since $-\frac{43}{2}$ and $\frac{4}{2}$ have the same denominator, add them by adding their numerators.

$$\frac{-43+4}{2}$$

Add $-43$ and $4$ to get $-39$.

$$-\frac{39}{2}$$

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