Solve 1/7+5/8div3/4

$$\frac { 1 } { 7 } + \frac { 5 } { 8 } \div \frac { 3 } { 4 }$$

Evaluate

$$\frac{41}{42}\approx 0.976190476$$

Solution Steps

Divide \frac{5}{8} by \frac{3}{4} by multiplying \frac{5}{8} by the reciprocal of \frac{3}{4}.

$$\frac{1}{7}+\frac{5}{8}\times \frac{4}{3}$$

Multiply \frac{5}{8} times \frac{4}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{7}+\frac{5\times 4}{8\times 3}$$

Do the multiplications in the fraction \frac{5\times 4}{8\times 3}.

$$\frac{1}{7}+\frac{20}{24}$$

Reduce the fraction \frac{20}{24} to lowest terms by extracting and canceling out 4.

$$\frac{1}{7}+\frac{5}{6}$$

Least common multiple of 7 and 6 is 42. Convert \frac{1}{7} and \frac{5}{6} to fractions with denominator 42.

$$\frac{6}{42}+\frac{35}{42}$$

Since \frac{6}{42} and \frac{35}{42} have the same denominator, add them by adding their numerators.

$$\frac{6+35}{42}$$

Add 6 and 35 to get 41.

$$\frac{41}{42}$$

Factor

$$\frac{41}{2 \cdot 3 \cdot 7} = 0.9761904761904762$$