Solve -7-11/3-(-7)-42/3

$$-7-1 \frac { 1 } { 3 } -(-7)- 4 \frac { 2 } { 3 }$$

Evaluate

$$-6$$

Solution Steps

Multiply 1 and 3 to get 3.

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

Add 3 and 1 to get 4.

$$-7-\frac{4}{3}-\left(-7\right)-\frac{4\times 3+2}{3}$$

Convert -7 to fraction -\frac{21}{3}.

$$-\frac{21}{3}-\frac{4}{3}-\left(-7\right)-\frac{4\times 3+2}{3}$$

Since -\frac{21}{3} and \frac{4}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-21-4}{3}-\left(-7\right)-\frac{4\times 3+2}{3}$$

Subtract 4 from -21 to get -25.

$$-\frac{25}{3}-\left(-7\right)-\frac{4\times 3+2}{3}$$

The opposite of -7 is 7.

$$-\frac{25}{3}+7-\frac{4\times 3+2}{3}$$

Convert 7 to fraction \frac{21}{3}.

$$-\frac{25}{3}+\frac{21}{3}-\frac{4\times 3+2}{3}$$

Since -\frac{25}{3} and \frac{21}{3} have the same denominator, add them by adding their numerators.

$$\frac{-25+21}{3}-\frac{4\times 3+2}{3}$$

Add -25 and 21 to get -4.

$$-\frac{4}{3}-\frac{4\times 3+2}{3}$$

Multiply 4 and 3 to get 12.

$$-\frac{4}{3}-\frac{12+2}{3}$$

Add 12 and 2 to get 14.

$$-\frac{4}{3}-\frac{14}{3}$$

Since -\frac{4}{3} and \frac{14}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-4-14}{3}$$

Subtract 14 from -4 to get -18.

$$\frac{-18}{3}$$

Divide -18 by 3 to get -6.

$$-6$$

Factor

$$-6$$