Solve (|3/2+2/2-7/3|+|2/3-3|)

$$( | \frac { 3 } { 2 } + \frac { 2 } { 2 } - \frac { 7 } { 3 } | + | \frac { 2 } { 3 } - 3 | )$$

Evaluate

$$\frac{5}{2}=2.5$$

Solution Steps

Since \frac{3}{2} and \frac{2}{2} have the same denominator, add them by adding their numerators.

$$|\frac{3+2}{2}-\frac{7}{3}|+|\frac{2}{3}-3|$$

Add 3 and 2 to get 5.

$$|\frac{5}{2}-\frac{7}{3}|+|\frac{2}{3}-3|$$

Least common multiple of 2 and 3 is 6. Convert \frac{5}{2} and \frac{7}{3} to fractions with denominator 6.

$$|\frac{15}{6}-\frac{14}{6}|+|\frac{2}{3}-3|$$

Since \frac{15}{6} and \frac{14}{6} have the same denominator, subtract them by subtracting their numerators.

$$|\frac{15-14}{6}|+|\frac{2}{3}-3|$$

Subtract 14 from 15 to get 1.

$$|\frac{1}{6}|+|\frac{2}{3}-3|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of \frac{1}{6} is \frac{1}{6}.

$$\frac{1}{6}+|\frac{2}{3}-3|$$

Convert 3 to fraction \frac{9}{3}.

$$\frac{1}{6}+|\frac{2}{3}-\frac{9}{3}|$$

Since \frac{2}{3} and \frac{9}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{6}+|\frac{2-9}{3}|$$

Subtract 9 from 2 to get -7.

$$\frac{1}{6}+|-\frac{7}{3}|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -\frac{7}{3} is \frac{7}{3}.

$$\frac{1}{6}+\frac{7}{3}$$

Least common multiple of 6 and 3 is 6. Convert \frac{1}{6} and \frac{7}{3} to fractions with denominator 6.

$$\frac{1}{6}+\frac{14}{6}$$

Since \frac{1}{6} and \frac{14}{6} have the same denominator, add them by adding their numerators.

$$\frac{1+14}{6}$$

Add 1 and 14 to get 15.

$$\frac{15}{6}$$

Reduce the fraction \frac{15}{6} to lowest terms by extracting and canceling out 3.

$$\frac{5}{2}$$

Factor

$$\frac{5}{2} = 2\frac{1}{2} = 2.5$$