Solve (3/4-1/3):(7/2+1/4)+(frac{7}{12}+frac{1}{3}):(frac{5}{4}+frac{1}{3})

$$( \frac { 3 } { 4 } - \frac { 1 } { 3 } ) : ( \frac { 7 } { 2 } + \frac { 1 } { 4 } ) + ( \frac { 7 } { 12 } + \frac { 1 } { 3 } ) : ( \frac { 5 } { 4 } + \frac { 1 } { 3 } )$$

Evaluate

$$\frac{118}{171}\approx 0.69005848$$

Solution Steps

Least common multiple of 4 and 3 is 12. Convert \frac{3}{4} and \frac{1}{3} to fractions with denominator 12.

$$\frac{\frac{9}{12}-\frac{4}{12}}{\frac{7}{2}+\frac{1}{4}}+\frac{\frac{7}{12}+\frac{1}{3}}{\frac{5}{4}+\frac{1}{3}}$$

Since \frac{9}{12} and \frac{4}{12} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{9-4}{12}}{\frac{7}{2}+\frac{1}{4}}+\frac{\frac{7}{12}+\frac{1}{3}}{\frac{5}{4}+\frac{1}{3}}$$

Subtract 4 from 9 to get 5.

$$\frac{\frac{5}{12}}{\frac{7}{2}+\frac{1}{4}}+\frac{\frac{7}{12}+\frac{1}{3}}{\frac{5}{4}+\frac{1}{3}}$$

Least common multiple of 2 and 4 is 4. Convert \frac{7}{2} and \frac{1}{4} to fractions with denominator 4.

$$\frac{\frac{5}{12}}{\frac{14}{4}+\frac{1}{4}}+\frac{\frac{7}{12}+\frac{1}{3}}{\frac{5}{4}+\frac{1}{3}}$$

Since \frac{14}{4} and \frac{1}{4} have the same denominator, add them by adding their numerators.

$$\frac{\frac{5}{12}}{\frac{14+1}{4}}+\frac{\frac{7}{12}+\frac{1}{3}}{\frac{5}{4}+\frac{1}{3}}$$

Add 14 and 1 to get 15.

$$\frac{\frac{5}{12}}{\frac{15}{4}}+\frac{\frac{7}{12}+\frac{1}{3}}{\frac{5}{4}+\frac{1}{3}}$$

Divide \frac{5}{12} by \frac{15}{4} by multiplying \frac{5}{12} by the reciprocal of \frac{15}{4}.

$$\frac{5}{12}\times \frac{4}{15}+\frac{\frac{7}{12}+\frac{1}{3}}{\frac{5}{4}+\frac{1}{3}}$$

Multiply \frac{5}{12} times \frac{4}{15} by multiplying numerator times numerator and denominator times denominator.

$$\frac{5\times 4}{12\times 15}+\frac{\frac{7}{12}+\frac{1}{3}}{\frac{5}{4}+\frac{1}{3}}$$

Do the multiplications in the fraction \frac{5\times 4}{12\times 15}.

$$\frac{20}{180}+\frac{\frac{7}{12}+\frac{1}{3}}{\frac{5}{4}+\frac{1}{3}}$$

Reduce the fraction \frac{20}{180} to lowest terms by extracting and canceling out 20.

$$\frac{1}{9}+\frac{\frac{7}{12}+\frac{1}{3}}{\frac{5}{4}+\frac{1}{3}}$$

Least common multiple of 12 and 3 is 12. Convert \frac{7}{12} and \frac{1}{3} to fractions with denominator 12.

$$\frac{1}{9}+\frac{\frac{7}{12}+\frac{4}{12}}{\frac{5}{4}+\frac{1}{3}}$$

Since \frac{7}{12} and \frac{4}{12} have the same denominator, add them by adding their numerators.

$$\frac{1}{9}+\frac{\frac{7+4}{12}}{\frac{5}{4}+\frac{1}{3}}$$

Add 7 and 4 to get 11.

$$\frac{1}{9}+\frac{\frac{11}{12}}{\frac{5}{4}+\frac{1}{3}}$$

Least common multiple of 4 and 3 is 12. Convert \frac{5}{4} and \frac{1}{3} to fractions with denominator 12.

$$\frac{1}{9}+\frac{\frac{11}{12}}{\frac{15}{12}+\frac{4}{12}}$$

Since \frac{15}{12} and \frac{4}{12} have the same denominator, add them by adding their numerators.

$$\frac{1}{9}+\frac{\frac{11}{12}}{\frac{15+4}{12}}$$

Add 15 and 4 to get 19.

$$\frac{1}{9}+\frac{\frac{11}{12}}{\frac{19}{12}}$$

Divide \frac{11}{12} by \frac{19}{12} by multiplying \frac{11}{12} by the reciprocal of \frac{19}{12}.

$$\frac{1}{9}+\frac{11}{12}\times \frac{12}{19}$$

Multiply \frac{11}{12} times \frac{12}{19} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{9}+\frac{11\times 12}{12\times 19}$$

Cancel out 12 in both numerator and denominator.

$$\frac{1}{9}+\frac{11}{19}$$

Least common multiple of 9 and 19 is 171. Convert \frac{1}{9} and \frac{11}{19} to fractions with denominator 171.

$$\frac{19}{171}+\frac{99}{171}$$

Since \frac{19}{171} and \frac{99}{171} have the same denominator, add them by adding their numerators.

$$\frac{19+99}{171}$$

Add 19 and 99 to get 118.

$$\frac{118}{171}$$

Factor

$$\frac{2 \cdot 59}{3 ^ {2} \cdot 19} = 0.6900584795321637$$