Solve 2/9(12/3*114/9)+frac{4/17}{frac{3}{4}*1frac{4}{7}:1frac{1}{2}-frac{11}{28}}*frac{frac{3}{4}+frac{2}{3}}{frac{3}{4}-frac{2}{3}}-21>0

$$\frac { \frac { 2 } { 9 } ( 1 \frac { 2 } { 3 } \times 11 \frac { 4 } { 9 } ) + \frac { 4 } { 17 } } { \frac { 3 } { 4 } \times 1 \frac { 4 } { 7 } : 1 \frac { 1 } { 2 } - \frac { 11 } { 28 } } \times \frac { \frac { 3 } { 4 } + \frac { 2 } { 3 } } { \frac { 3 } { 4 } - \frac { 2 } { 3 } } - 21 > 0$$

Verify

Solution Steps

Multiply 1 and 3 to get 3.

$$\frac{\frac{2}{9}\times \frac{3+2}{3}\times \frac{11\times 9+4}{9}+\frac{4}{17}}{\frac{\frac{3}{4}\times \frac{1\times 7+4}{7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Add 3 and 2 to get 5.

$$\frac{\frac{2}{9}\times \frac{5}{3}\times \frac{11\times 9+4}{9}+\frac{4}{17}}{\frac{\frac{3}{4}\times \frac{1\times 7+4}{7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Multiply \frac{2}{9} times \frac{5}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{2\times 5}{9\times 3}\times \frac{11\times 9+4}{9}+\frac{4}{17}}{\frac{\frac{3}{4}\times \frac{1\times 7+4}{7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Do the multiplications in the fraction \frac{2\times 5}{9\times 3}.

$$\frac{\frac{10}{27}\times \frac{11\times 9+4}{9}+\frac{4}{17}}{\frac{\frac{3}{4}\times \frac{1\times 7+4}{7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Multiply 11 and 9 to get 99.

$$\frac{\frac{10}{27}\times \frac{99+4}{9}+\frac{4}{17}}{\frac{\frac{3}{4}\times \frac{1\times 7+4}{7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Add 99 and 4 to get 103.

$$\frac{\frac{10}{27}\times \frac{103}{9}+\frac{4}{17}}{\frac{\frac{3}{4}\times \frac{1\times 7+4}{7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Multiply \frac{10}{27} times \frac{103}{9} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{10\times 103}{27\times 9}+\frac{4}{17}}{\frac{\frac{3}{4}\times \frac{1\times 7+4}{7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Do the multiplications in the fraction \frac{10\times 103}{27\times 9}.

$$\frac{\frac{1030}{243}+\frac{4}{17}}{\frac{\frac{3}{4}\times \frac{1\times 7+4}{7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Least common multiple of 243 and 17 is 4131. Convert \frac{1030}{243} and \frac{4}{17} to fractions with denominator 4131.

$$\frac{\frac{17510}{4131}+\frac{972}{4131}}{\frac{\frac{3}{4}\times \frac{1\times 7+4}{7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Since \frac{17510}{4131} and \frac{972}{4131} have the same denominator, add them by adding their numerators.

$$\frac{\frac{17510+972}{4131}}{\frac{\frac{3}{4}\times \frac{1\times 7+4}{7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Add 17510 and 972 to get 18482.

$$\frac{\frac{18482}{4131}}{\frac{\frac{3}{4}\times \frac{1\times 7+4}{7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Multiply 1 and 7 to get 7.

$$\frac{\frac{18482}{4131}}{\frac{\frac{3}{4}\times \frac{7+4}{7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Add 7 and 4 to get 11.

$$\frac{\frac{18482}{4131}}{\frac{\frac{3}{4}\times \frac{11}{7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Multiply \frac{3}{4} times \frac{11}{7} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{18482}{4131}}{\frac{\frac{3\times 11}{4\times 7}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Do the multiplications in the fraction \frac{3\times 11}{4\times 7}.

$$\frac{\frac{18482}{4131}}{\frac{\frac{33}{28}}{\frac{1\times 2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Multiply 1 and 2 to get 2.

$$\frac{\frac{18482}{4131}}{\frac{\frac{33}{28}}{\frac{2+1}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Add 2 and 1 to get 3.

$$\frac{\frac{18482}{4131}}{\frac{\frac{33}{28}}{\frac{3}{2}}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Divide \frac{33}{28} by \frac{3}{2} by multiplying \frac{33}{28} by the reciprocal of \frac{3}{2}.

$$\frac{\frac{18482}{4131}}{\frac{33}{28}\times \frac{2}{3}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Multiply \frac{33}{28} times \frac{2}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{18482}{4131}}{\frac{33\times 2}{28\times 3}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Do the multiplications in the fraction \frac{33\times 2}{28\times 3}.

$$\frac{\frac{18482}{4131}}{\frac{66}{84}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Reduce the fraction \frac{66}{84} to lowest terms by extracting and canceling out 6.

$$\frac{\frac{18482}{4131}}{\frac{11}{14}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Least common multiple of 14 and 28 is 28. Convert \frac{11}{14} and \frac{11}{28} to fractions with denominator 28.

$$\frac{\frac{18482}{4131}}{\frac{22}{28}-\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Since \frac{22}{28} and \frac{11}{28} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{18482}{4131}}{\frac{22-11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Subtract 11 from 22 to get 11.

$$\frac{\frac{18482}{4131}}{\frac{11}{28}}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Divide \frac{18482}{4131} by \frac{11}{28} by multiplying \frac{18482}{4131} by the reciprocal of \frac{11}{28}.

$$\frac{18482}{4131}\times \frac{28}{11}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Multiply \frac{18482}{4131} times \frac{28}{11} by multiplying numerator times numerator and denominator times denominator.

$$\frac{18482\times 28}{4131\times 11}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Do the multiplications in the fraction \frac{18482\times 28}{4131\times 11}.

$$\frac{517496}{45441}\times \frac{\frac{3}{4}+\frac{2}{3}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Least common multiple of 4 and 3 is 12. Convert \frac{3}{4} and \frac{2}{3} to fractions with denominator 12.

$$\frac{517496}{45441}\times \frac{\frac{9}{12}+\frac{8}{12}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Since \frac{9}{12} and \frac{8}{12} have the same denominator, add them by adding their numerators.

$$\frac{517496}{45441}\times \frac{\frac{9+8}{12}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Add 9 and 8 to get 17.

$$\frac{517496}{45441}\times \frac{\frac{17}{12}}{\frac{3}{4}-\frac{2}{3}}-21>0$$

Least common multiple of 4 and 3 is 12. Convert \frac{3}{4} and \frac{2}{3} to fractions with denominator 12.

$$\frac{517496}{45441}\times \frac{\frac{17}{12}}{\frac{9}{12}-\frac{8}{12}}-21>0$$

Since \frac{9}{12} and \frac{8}{12} have the same denominator, subtract them by subtracting their numerators.

$$\frac{517496}{45441}\times \frac{\frac{17}{12}}{\frac{9-8}{12}}-21>0$$

Subtract 8 from 9 to get 1.

$$\frac{517496}{45441}\times \frac{\frac{17}{12}}{\frac{1}{12}}-21>0$$

Divide \frac{17}{12} by \frac{1}{12} by multiplying \frac{17}{12} by the reciprocal of \frac{1}{12}.

$$\frac{517496}{45441}\times \frac{17}{12}\times 12-21>0$$

Cancel out 12 and 12.

$$\frac{517496}{45441}\times 17-21>0$$

Express \frac{517496}{45441}\times 17 as a single fraction.

$$\frac{517496\times 17}{45441}-21>0$$

Multiply 517496 and 17 to get 8797432.

$$\frac{8797432}{45441}-21>0$$

Reduce the fraction \frac{8797432}{45441} to lowest terms by extracting and canceling out 17.

$$\frac{517496}{2673}-21>0$$

Convert 21 to fraction \frac{56133}{2673}.

$$\frac{517496}{2673}-\frac{56133}{2673}>0$$

Since \frac{517496}{2673} and \frac{56133}{2673} have the same denominator, subtract them by subtracting their numerators.

$$\frac{517496-56133}{2673}>0$$

Subtract 56133 from 517496 to get 461363.

$$\frac{461363}{2673}>0$$

Compare \frac{461363}{2673} and 0.

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