Solve (2+2/7)(1+7/6)(2+2/3)*1/8=frac{3}{4}

$$( 2 + \frac { 2 } { 7 } ) ( 1 + \frac { 7 } { 6 } ) ( 2 + \frac { 2 } { 3 } ) \cdot \frac { 1 } { 8 } = \frac { 3 } { 4 }$$

Verify

Solution Steps

Convert 2 to fraction \frac{14}{7}.

$$\left(\frac{14}{7}+\frac{2}{7}\right)\left(1+\frac{7}{6}\right)\left(2+\frac{2}{3}\right)\times \frac{1}{8}=\frac{3}{4}$$

Since \frac{14}{7} and \frac{2}{7} have the same denominator, add them by adding their numerators.

$$\frac{14+2}{7}\left(1+\frac{7}{6}\right)\left(2+\frac{2}{3}\right)\times \frac{1}{8}=\frac{3}{4}$$

Add 14 and 2 to get 16.

$$\frac{16}{7}\left(1+\frac{7}{6}\right)\left(2+\frac{2}{3}\right)\times \frac{1}{8}=\frac{3}{4}$$

Convert 1 to fraction \frac{6}{6}.

$$\frac{16}{7}\left(\frac{6}{6}+\frac{7}{6}\right)\left(2+\frac{2}{3}\right)\times \frac{1}{8}=\frac{3}{4}$$

Since \frac{6}{6} and \frac{7}{6} have the same denominator, add them by adding their numerators.

$$\frac{16}{7}\times \frac{6+7}{6}\left(2+\frac{2}{3}\right)\times \frac{1}{8}=\frac{3}{4}$$

Add 6 and 7 to get 13.

$$\frac{16}{7}\times \frac{13}{6}\left(2+\frac{2}{3}\right)\times \frac{1}{8}=\frac{3}{4}$$

Multiply \frac{16}{7} times \frac{13}{6} by multiplying numerator times numerator and denominator times denominator.

$$\frac{16\times 13}{7\times 6}\left(2+\frac{2}{3}\right)\times \frac{1}{8}=\frac{3}{4}$$

Do the multiplications in the fraction \frac{16\times 13}{7\times 6}.

$$\frac{208}{42}\left(2+\frac{2}{3}\right)\times \frac{1}{8}=\frac{3}{4}$$

Reduce the fraction \frac{208}{42} to lowest terms by extracting and canceling out 2.

$$\frac{104}{21}\left(2+\frac{2}{3}\right)\times \frac{1}{8}=\frac{3}{4}$$

Convert 2 to fraction \frac{6}{3}.

$$\frac{104}{21}\left(\frac{6}{3}+\frac{2}{3}\right)\times \frac{1}{8}=\frac{3}{4}$$

Since \frac{6}{3} and \frac{2}{3} have the same denominator, add them by adding their numerators.

$$\frac{104}{21}\times \frac{6+2}{3}\times \frac{1}{8}=\frac{3}{4}$$

Add 6 and 2 to get 8.

$$\frac{104}{21}\times \frac{8}{3}\times \frac{1}{8}=\frac{3}{4}$$

Multiply \frac{104}{21} times \frac{8}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{104\times 8}{21\times 3}\times \frac{1}{8}=\frac{3}{4}$$

Do the multiplications in the fraction \frac{104\times 8}{21\times 3}.

$$\frac{832}{63}\times \frac{1}{8}=\frac{3}{4}$$

Multiply \frac{832}{63} times \frac{1}{8} by multiplying numerator times numerator and denominator times denominator.

$$\frac{832\times 1}{63\times 8}=\frac{3}{4}$$

Do the multiplications in the fraction \frac{832\times 1}{63\times 8}.

$$\frac{832}{504}=\frac{3}{4}$$

Reduce the fraction \frac{832}{504} to lowest terms by extracting and canceling out 8.

$$\frac{104}{63}=\frac{3}{4}$$

Least common multiple of 63 and 4 is 252. Convert \frac{104}{63} and \frac{3}{4} to fractions with denominator 252.

$$\frac{416}{252}=\frac{189}{252}$$

Compare \frac{416}{252} and \frac{189}{252}.

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