Solve (11/49-3/7)div(11/49+3/7)

$$( \frac { 11 } { 49 } - \frac { 3 } { 7 } ) \div ( \frac { 11 } { 49 } + \frac { 3 } { 7 } )$$

Evaluate

$$-\frac{5}{16}=-0.3125$$

Solution Steps

Least common multiple of 49 and 7 is 49. Convert \frac{11}{49} and \frac{3}{7} to fractions with denominator 49.

$$\frac{\frac{11}{49}-\frac{21}{49}}{\frac{11}{49}+\frac{3}{7}}$$

Since \frac{11}{49} and \frac{21}{49} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{11-21}{49}}{\frac{11}{49}+\frac{3}{7}}$$

Subtract 21 from 11 to get -10.

$$\frac{-\frac{10}{49}}{\frac{11}{49}+\frac{3}{7}}$$

Least common multiple of 49 and 7 is 49. Convert \frac{11}{49} and \frac{3}{7} to fractions with denominator 49.

$$\frac{-\frac{10}{49}}{\frac{11}{49}+\frac{21}{49}}$$

Since \frac{11}{49} and \frac{21}{49} have the same denominator, add them by adding their numerators.

$$\frac{-\frac{10}{49}}{\frac{11+21}{49}}$$

Add 11 and 21 to get 32.

$$\frac{-\frac{10}{49}}{\frac{32}{49}}$$

Divide -\frac{10}{49} by \frac{32}{49} by multiplying -\frac{10}{49} by the reciprocal of \frac{32}{49}.

$$-\frac{10}{49}\times \frac{49}{32}$$

Multiply -\frac{10}{49} times \frac{49}{32} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-10\times 49}{49\times 32}$$

Cancel out 49 in both numerator and denominator.

$$\frac{-10}{32}$$

Reduce the fraction \frac{-10}{32} to lowest terms by extracting and canceling out 2.

$$-\frac{5}{16}$$

Factor

$$-\frac{5}{16} = -0.3125$$