Solve [(7-1/3)-1]:{[(7/15-3/8*4/5)+frac{1}{3}]*frac{9}{4}+1}=(frac{5}{3}+1)

$$[ ( 7 - \frac { 1 } { 3 } ) - 1 ] : \{ [ ( \frac { 7 } { 15 } - \frac { 3 } { 8 } \cdot \frac { 4 } { 5 } ) + \frac { 1 } { 3 } ] \cdot \frac { 9 } { 4 } + 1 \} = ( \frac { 5 } { 3 } + 1 )$$

Verify

Solution Steps

Convert 7 to fraction \frac{21}{3}.

$$\frac{\frac{21}{3}-\frac{1}{3}-1}{\left(\frac{7}{15}-\frac{3}{8}\times \frac{4}{5}+\frac{1}{3}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Since \frac{21}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{21-1}{3}-1}{\left(\frac{7}{15}-\frac{3}{8}\times \frac{4}{5}+\frac{1}{3}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Subtract 1 from 21 to get 20.

$$\frac{\frac{20}{3}-1}{\left(\frac{7}{15}-\frac{3}{8}\times \frac{4}{5}+\frac{1}{3}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Convert 1 to fraction \frac{3}{3}.

$$\frac{\frac{20}{3}-\frac{3}{3}}{\left(\frac{7}{15}-\frac{3}{8}\times \frac{4}{5}+\frac{1}{3}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Since \frac{20}{3} and \frac{3}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{20-3}{3}}{\left(\frac{7}{15}-\frac{3}{8}\times \frac{4}{5}+\frac{1}{3}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Subtract 3 from 20 to get 17.

$$\frac{\frac{17}{3}}{\left(\frac{7}{15}-\frac{3}{8}\times \frac{4}{5}+\frac{1}{3}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

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

$$\frac{\frac{17}{3}}{\left(\frac{7}{15}-\frac{3\times 4}{8\times 5}+\frac{1}{3}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

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

$$\frac{\frac{17}{3}}{\left(\frac{7}{15}-\frac{12}{40}+\frac{1}{3}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Reduce the fraction \frac{12}{40} to lowest terms by extracting and canceling out 4.

$$\frac{\frac{17}{3}}{\left(\frac{7}{15}-\frac{3}{10}+\frac{1}{3}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Least common multiple of 15 and 10 is 30. Convert \frac{7}{15} and \frac{3}{10} to fractions with denominator 30.

$$\frac{\frac{17}{3}}{\left(\frac{14}{30}-\frac{9}{30}+\frac{1}{3}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Since \frac{14}{30} and \frac{9}{30} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{17}{3}}{\left(\frac{14-9}{30}+\frac{1}{3}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Subtract 9 from 14 to get 5.

$$\frac{\frac{17}{3}}{\left(\frac{5}{30}+\frac{1}{3}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Reduce the fraction \frac{5}{30} to lowest terms by extracting and canceling out 5.

$$\frac{\frac{17}{3}}{\left(\frac{1}{6}+\frac{1}{3}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Least common multiple of 6 and 3 is 6. Convert \frac{1}{6} and \frac{1}{3} to fractions with denominator 6.

$$\frac{\frac{17}{3}}{\left(\frac{1}{6}+\frac{2}{6}\right)\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Since \frac{1}{6} and \frac{2}{6} have the same denominator, add them by adding their numerators.

$$\frac{\frac{17}{3}}{\frac{1+2}{6}\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Add 1 and 2 to get 3.

$$\frac{\frac{17}{3}}{\frac{3}{6}\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

$$\frac{\frac{17}{3}}{\frac{1}{2}\times \frac{9}{4}+1}=\frac{5}{3}+1$$

Multiply \frac{1}{2} times \frac{9}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{17}{3}}{\frac{1\times 9}{2\times 4}+1}=\frac{5}{3}+1$$

Do the multiplications in the fraction \frac{1\times 9}{2\times 4}.

$$\frac{\frac{17}{3}}{\frac{9}{8}+1}=\frac{5}{3}+1$$

Convert 1 to fraction \frac{8}{8}.

$$\frac{\frac{17}{3}}{\frac{9}{8}+\frac{8}{8}}=\frac{5}{3}+1$$

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

$$\frac{\frac{17}{3}}{\frac{9+8}{8}}=\frac{5}{3}+1$$

Add 9 and 8 to get 17.

$$\frac{\frac{17}{3}}{\frac{17}{8}}=\frac{5}{3}+1$$

Divide \frac{17}{3} by \frac{17}{8} by multiplying \frac{17}{3} by the reciprocal of \frac{17}{8}.

$$\frac{17}{3}\times \frac{8}{17}=\frac{5}{3}+1$$

Multiply \frac{17}{3} times \frac{8}{17} by multiplying numerator times numerator and denominator times denominator.

$$\frac{17\times 8}{3\times 17}=\frac{5}{3}+1$$

Cancel out 17 in both numerator and denominator.

$$\frac{8}{3}=\frac{5}{3}+1$$

Convert 1 to fraction \frac{3}{3}.

$$\frac{8}{3}=\frac{5}{3}+\frac{3}{3}$$

Since \frac{5}{3} and \frac{3}{3} have the same denominator, add them by adding their numerators.

$$\frac{8}{3}=\frac{5+3}{3}$$

Add 5 and 3 to get 8.

$$\frac{8}{3}=\frac{8}{3}$$

Compare \frac{8}{3} and \frac{8}{3}.

$$\text{true}$$