Solve 63/4-42/3+7/12

$$6 \frac { 3 } { 4 } - \frac { 42 } { 3 } + \frac { 7 } { 12 }$$

Evaluate

$$-\frac{20}{3}\approx -6.666666667$$

Solution Steps

Multiply 6 and 4 to get 24.

$$\frac{24+3}{4}-\frac{42}{3}+\frac{7}{12}$$

Add 24 and 3 to get 27.

$$\frac{27}{4}-\frac{42}{3}+\frac{7}{12}$$

Divide 42 by 3 to get 14.

$$\frac{27}{4}-14+\frac{7}{12}$$

Convert 14 to fraction \frac{56}{4}.

$$\frac{27}{4}-\frac{56}{4}+\frac{7}{12}$$

Since \frac{27}{4} and \frac{56}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{27-56}{4}+\frac{7}{12}$$

Subtract 56 from 27 to get -29.

$$-\frac{29}{4}+\frac{7}{12}$$

Least common multiple of 4 and 12 is 12. Convert -\frac{29}{4} and \frac{7}{12} to fractions with denominator 12.

$$-\frac{87}{12}+\frac{7}{12}$$

Since -\frac{87}{12} and \frac{7}{12} have the same denominator, add them by adding their numerators.

$$\frac{-87+7}{12}$$

Add -87 and 7 to get -80.

$$\frac{-80}{12}$$

Reduce the fraction \frac{-80}{12} to lowest terms by extracting and canceling out 4.

$$-\frac{20}{3}$$

Factor

$$-\frac{20}{3} = -6\frac{2}{3} = -6.666666666666667$$