Solve -9*1^2/3+6*1/3+3=

$$- 9 \cdot \frac { 1 ^ { 2 } } { 3 } + 6 \cdot \frac { 1 } { 3 } + 3 =$$

Evaluate

$$2$$

Solution Steps

Calculate 1 to the power of 2 and get 1.

$$-9\times \frac{1}{3}+6\times \frac{1}{3}+3$$

Multiply -9 and \frac{1}{3} to get \frac{-9}{3}.

$$\frac{-9}{3}+6\times \frac{1}{3}+3$$

Divide -9 by 3 to get -3.

$$-3+6\times \frac{1}{3}+3$$

Multiply 6 and \frac{1}{3} to get \frac{6}{3}.

$$-3+\frac{6}{3}+3$$

Divide 6 by 3 to get 2.

$$-3+2+3$$

Add -3 and 2 to get -1.

$$-1+3$$

Add -1 and 3 to get 2.

$$2$$

Factor

$$2$$