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\frac{\frac{1\times 3+2}{3}-\frac{1}{3}}{1\times \frac{7\times 2+2}{2}}
Divide 7 by 7 to get 1.
\frac{\frac{3+2}{3}-\frac{1}{3}}{1\times \frac{7\times 2+2}{2}}
Multiply 1 and 3 to get 3.
\frac{\frac{5}{3}-\frac{1}{3}}{1\times \frac{7\times 2+2}{2}}
Add 3 and 2 to get 5.
\frac{\frac{5-1}{3}}{1\times \frac{7\times 2+2}{2}}
Since \frac{5}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.
\frac{\frac{4}{3}}{1\times \frac{7\times 2+2}{2}}
Subtract 1 from 5 to get 4.
\frac{\frac{4}{3}}{1\times \frac{14+2}{2}}
Multiply 7 and 2 to get 14.
\frac{\frac{4}{3}}{1\times \frac{16}{2}}
Add 14 and 2 to get 16.
\frac{\frac{4}{3}}{1\times 8}
Divide 16 by 2 to get 8.
\frac{\frac{4}{3}}{8}
Multiply 1 and 8 to get 8.
\frac{4}{3\times 8}
Express \frac{\frac{4}{3}}{8} as a single fraction.
\frac{4}{24}
Multiply 3 and 8 to get 24.
\frac{1}{6}
Reduce the fraction \frac{4}{24} to lowest terms by extracting and canceling out 4.