Solve -96/3quad+3/3-4/3

$$- \frac { 96 } { 3 } \quad + \frac { 3 } { 3 } - \frac { 4 } { 3 }$$

Evaluate

$$-\frac{97}{3}\approx -32.333333333$$

Solution Steps

Divide 3 by 3 to get 1.

$$-\frac{96}{3}+1-\frac{4}{3}$$

Divide 96 by 3 to get 32.

$$-32+1-\frac{4}{3}$$

Add -32 and 1 to get -31.

$$-31-\frac{4}{3}$$

Convert -31 to fraction -\frac{93}{3}.

$$-\frac{93}{3}-\frac{4}{3}$$

Since -\frac{93}{3} and \frac{4}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-93-4}{3}$$

Subtract 4 from -93 to get -97.

$$-\frac{97}{3}$$

Factor

$$-\frac{97}{3} = -32\frac{1}{3} = -32.333333333333336$$