Solve 10-(1-1/3+21/2)=

$$10 - ( 1 - \frac { 1 } { 3 } + 2 \frac { 1 } { 2 } ) =$$

Evaluate

$$\frac{41}{6}\approx 6.833333333$$

Solution Steps

Convert 1 to fraction \frac{3}{3}.

$$10-\left(\frac{3}{3}-\frac{1}{3}+\frac{2\times 2+1}{2}\right)$$

Since \frac{3}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

$$10-\left(\frac{3-1}{3}+\frac{2\times 2+1}{2}\right)$$

Subtract 1 from 3 to get 2.

$$10-\left(\frac{2}{3}+\frac{2\times 2+1}{2}\right)$$

Multiply 2 and 2 to get 4.

$$10-\left(\frac{2}{3}+\frac{4+1}{2}\right)$$

Add 4 and 1 to get 5.

$$10-\left(\frac{2}{3}+\frac{5}{2}\right)$$

Least common multiple of 3 and 2 is 6. Convert \frac{2}{3} and \frac{5}{2} to fractions with denominator 6.

$$10-\left(\frac{4}{6}+\frac{15}{6}\right)$$

Since \frac{4}{6} and \frac{15}{6} have the same denominator, add them by adding their numerators.

$$10-\frac{4+15}{6}$$

Add 4 and 15 to get 19.

$$10-\frac{19}{6}$$

Convert 10 to fraction \frac{60}{6}.

$$\frac{60}{6}-\frac{19}{6}$$

Since \frac{60}{6} and \frac{19}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{60-19}{6}$$

Subtract 19 from 60 to get 41.

$$\frac{41}{6}$$

Factor

$$\frac{41}{2 \cdot 3} = 6\frac{5}{6} = 6.833333333333333$$