Solve 2+3:(6+3)-7+(12*2)=

$$2 + 3 : ( 6 + 3 ) - 7 + ( 12 \times 2 ) =$$

Evaluate

$$\frac{58}{3}\approx 19.333333333$$

Solution Steps

Add 6 and 3 to get 9.

$$2+\frac{3}{9}-7+12\times 2$$

Reduce the fraction \frac{3}{9} to lowest terms by extracting and canceling out 3.

$$2+\frac{1}{3}-7+12\times 2$$

Convert 2 to fraction \frac{6}{3}.

$$\frac{6}{3}+\frac{1}{3}-7+12\times 2$$

Since \frac{6}{3} and \frac{1}{3} have the same denominator, add them by adding their numerators.

$$\frac{6+1}{3}-7+12\times 2$$

Add 6 and 1 to get 7.

$$\frac{7}{3}-7+12\times 2$$

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

$$\frac{7}{3}-\frac{21}{3}+12\times 2$$

Since \frac{7}{3} and \frac{21}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7-21}{3}+12\times 2$$

Subtract 21 from 7 to get -14.

$$-\frac{14}{3}+12\times 2$$

Multiply 12 and 2 to get 24.

$$-\frac{14}{3}+24$$

Convert 24 to fraction \frac{72}{3}.

$$-\frac{14}{3}+\frac{72}{3}$$

Since -\frac{14}{3} and \frac{72}{3} have the same denominator, add them by adding their numerators.

$$\frac{-14+72}{3}$$

Add -14 and 72 to get 58.

$$\frac{58}{3}$$

Factor

$$\frac{2 \cdot 29}{3} = 19\frac{1}{3} = 19.333333333333332$$