Solve {l}{13/4*8/11+(25/12-7/24)*3}{(3frac{1}{3}+frac{7}{12}):4+frac{2}{7}*frac{21}{48}}

$$\left. \begin{array} { l } { 1 \frac { 3 } { 4 } \cdot \frac { 8 } { 11 } + ( 2 \frac { 5 } { 12 } - \frac { 7 } { 24 } ) \cdot 3 } \\ { ( 3 \frac { 1 } { 3 } + \frac { 7 } { 12 } ) : 4 + \frac { 2 } { 7 } \cdot \frac { 21 } { 48 } } \end{array} \right.$$

Sort

$$\frac{53}{48},\frac{673}{88}$$

Solution Steps

Multiply 1 and 4 to get 4.

$$sort(\frac{4+3}{4}\times \frac{8}{11}+\left(\frac{2\times 12+5}{12}-\frac{7}{24}\right)\times 3,\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Add 4 and 3 to get 7.

$$sort(\frac{7}{4}\times \frac{8}{11}+\left(\frac{2\times 12+5}{12}-\frac{7}{24}\right)\times 3,\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Multiply \frac{7}{4} times \frac{8}{11} by multiplying numerator times numerator and denominator times denominator.

$$sort(\frac{7\times 8}{4\times 11}+\left(\frac{2\times 12+5}{12}-\frac{7}{24}\right)\times 3,\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Do the multiplications in the fraction \frac{7\times 8}{4\times 11}.

$$sort(\frac{56}{44}+\left(\frac{2\times 12+5}{12}-\frac{7}{24}\right)\times 3,\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Reduce the fraction \frac{56}{44} to lowest terms by extracting and canceling out 4.

$$sort(\frac{14}{11}+\left(\frac{2\times 12+5}{12}-\frac{7}{24}\right)\times 3,\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Multiply 2 and 12 to get 24.

$$sort(\frac{14}{11}+\left(\frac{24+5}{12}-\frac{7}{24}\right)\times 3,\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Add 24 and 5 to get 29.

$$sort(\frac{14}{11}+\left(\frac{29}{12}-\frac{7}{24}\right)\times 3,\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Least common multiple of 12 and 24 is 24. Convert \frac{29}{12} and \frac{7}{24} to fractions with denominator 24.

$$sort(\frac{14}{11}+\left(\frac{58}{24}-\frac{7}{24}\right)\times 3,\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Since \frac{58}{24} and \frac{7}{24} have the same denominator, subtract them by subtracting their numerators.

$$sort(\frac{14}{11}+\frac{58-7}{24}\times 3,\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Subtract 7 from 58 to get 51.

$$sort(\frac{14}{11}+\frac{51}{24}\times 3,\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Reduce the fraction \frac{51}{24} to lowest terms by extracting and canceling out 3.

$$sort(\frac{14}{11}+\frac{17}{8}\times 3,\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Express \frac{17}{8}\times 3 as a single fraction.

$$sort(\frac{14}{11}+\frac{17\times 3}{8},\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Multiply 17 and 3 to get 51.

$$sort(\frac{14}{11}+\frac{51}{8},\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Least common multiple of 11 and 8 is 88. Convert \frac{14}{11} and \frac{51}{8} to fractions with denominator 88.

$$sort(\frac{112}{88}+\frac{561}{88},\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Since \frac{112}{88} and \frac{561}{88} have the same denominator, add them by adding their numerators.

$$sort(\frac{112+561}{88},\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Add 112 and 561 to get 673.

$$sort(\frac{673}{88},\frac{\frac{3\times 3+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Multiply 3 and 3 to get 9.

$$sort(\frac{673}{88},\frac{\frac{9+1}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Add 9 and 1 to get 10.

$$sort(\frac{673}{88},\frac{\frac{10}{3}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Least common multiple of 3 and 12 is 12. Convert \frac{10}{3} and \frac{7}{12} to fractions with denominator 12.

$$sort(\frac{673}{88},\frac{\frac{40}{12}+\frac{7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Since \frac{40}{12} and \frac{7}{12} have the same denominator, add them by adding their numerators.

$$sort(\frac{673}{88},\frac{\frac{40+7}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Add 40 and 7 to get 47.

$$sort(\frac{673}{88},\frac{\frac{47}{12}}{4}+\frac{2}{7}\times \frac{21}{48})$$

Express \frac{\frac{47}{12}}{4} as a single fraction.

$$sort(\frac{673}{88},\frac{47}{12\times 4}+\frac{2}{7}\times \frac{21}{48})$$

Multiply 12 and 4 to get 48.

$$sort(\frac{673}{88},\frac{47}{48}+\frac{2}{7}\times \frac{21}{48})$$

Reduce the fraction \frac{21}{48} to lowest terms by extracting and canceling out 3.

$$sort(\frac{673}{88},\frac{47}{48}+\frac{2}{7}\times \frac{7}{16})$$

Multiply \frac{2}{7} times \frac{7}{16} by multiplying numerator times numerator and denominator times denominator.

$$sort(\frac{673}{88},\frac{47}{48}+\frac{2\times 7}{7\times 16})$$

Cancel out 7 in both numerator and denominator.

$$sort(\frac{673}{88},\frac{47}{48}+\frac{2}{16})$$

Reduce the fraction \frac{2}{16} to lowest terms by extracting and canceling out 2.

$$sort(\frac{673}{88},\frac{47}{48}+\frac{1}{8})$$

Least common multiple of 48 and 8 is 48. Convert \frac{47}{48} and \frac{1}{8} to fractions with denominator 48.

$$sort(\frac{673}{88},\frac{47}{48}+\frac{6}{48})$$

Since \frac{47}{48} and \frac{6}{48} have the same denominator, add them by adding their numerators.

$$sort(\frac{673}{88},\frac{47+6}{48})$$

Add 47 and 6 to get 53.

$$sort(\frac{673}{88},\frac{53}{48})$$

Least common denominator of the numbers in the list \frac{673}{88},\frac{53}{48} is 528. Convert numbers in the list to fractions with denominator 528.

$$\frac{4038}{528},\frac{583}{528}$$

To sort the list, start from a single element \frac{4038}{528}.

$$\frac{4038}{528}$$

Insert \frac{583}{528} to the appropriate location in the new list.

$$\frac{583}{528},\frac{4038}{528}$$

Replace the obtained fractions with the initial values.

$$\frac{53}{48},\frac{673}{88}$$

Evaluate

$$\frac{673}{88},\ \frac{53}{48}$$