Solve 2^2(3+4)-1^2(1/2+2/3)+20

$$2 ^ { 2 } ( 3 + 4 ) - 1 ^ { 2 } ( \frac { 1 } { 2 } + \frac { 2 } { 3 } ) + 20$$

Evaluate

$$\frac{281}{6}\approx 46.833333333$$

Solution Steps

Calculate 2 to the power of 2 and get 4.

$$4\left(3+4\right)-1^{2}\left(\frac{1}{2}+\frac{2}{3}\right)+20$$

Add 3 and 4 to get 7.

$$4\times 7-1^{2}\left(\frac{1}{2}+\frac{2}{3}\right)+20$$

Multiply 4 and 7 to get 28.

$$28-1^{2}\left(\frac{1}{2}+\frac{2}{3}\right)+20$$

Calculate 1 to the power of 2 and get 1.

$$28-1\left(\frac{1}{2}+\frac{2}{3}\right)+20$$

Least common multiple of 2 and 3 is 6. Convert \frac{1}{2} and \frac{2}{3} to fractions with denominator 6.

$$28-1\left(\frac{3}{6}+\frac{4}{6}\right)+20$$

Since \frac{3}{6} and \frac{4}{6} have the same denominator, add them by adding their numerators.

$$28-1\times \frac{3+4}{6}+20$$

Add 3 and 4 to get 7.

$$28-1\times \frac{7}{6}+20$$

Multiply 1 and \frac{7}{6} to get \frac{7}{6}.

$$28-\frac{7}{6}+20$$

Convert 28 to fraction \frac{168}{6}.

$$\frac{168}{6}-\frac{7}{6}+20$$

Since \frac{168}{6} and \frac{7}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{168-7}{6}+20$$

Subtract 7 from 168 to get 161.

$$\frac{161}{6}+20$$

Convert 20 to fraction \frac{120}{6}.

$$\frac{161}{6}+\frac{120}{6}$$

Since \frac{161}{6} and \frac{120}{6} have the same denominator, add them by adding their numerators.

$$\frac{161+120}{6}$$

Add 161 and 120 to get 281.

$$\frac{281}{6}$$

Factor

$$\frac{281}{2 \cdot 3} = 46\frac{5}{6} = 46.833333333333336$$