Solve (-7/2)*(1/6-1/2)*3/14=(-frac{1}{2})

$$( - \frac { 7 } { 2 } ) \times ( \frac { 1 } { 6 } - \frac { 1 } { 2 } ) \times \frac { 3 } { 14 } = ( - \frac { 1 } { 2 } )$$

Verify

Solution Steps

Least common multiple of 6 and 2 is 6. Convert \frac{1}{6} and \frac{1}{2} to fractions with denominator 6.

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

Since \frac{1}{6} and \frac{3}{6} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{7}{2}\times \frac{1-3}{6}\times \frac{3}{14}=-\frac{1}{2}$$

Subtract 3 from 1 to get -2.

$$-\frac{7}{2}\times \frac{-2}{6}\times \frac{3}{14}=-\frac{1}{2}$$

Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.

$$-\frac{7}{2}\left(-\frac{1}{3}\right)\times \frac{3}{14}=-\frac{1}{2}$$

Multiply -\frac{7}{2} times -\frac{1}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-7\left(-1\right)}{2\times 3}\times \frac{3}{14}=-\frac{1}{2}$$

Do the multiplications in the fraction \frac{-7\left(-1\right)}{2\times 3}.

$$\frac{7}{6}\times \frac{3}{14}=-\frac{1}{2}$$

Multiply \frac{7}{6} times \frac{3}{14} by multiplying numerator times numerator and denominator times denominator.

$$\frac{7\times 3}{6\times 14}=-\frac{1}{2}$$

Do the multiplications in the fraction \frac{7\times 3}{6\times 14}.

$$\frac{21}{84}=-\frac{1}{2}$$

Reduce the fraction \frac{21}{84} to lowest terms by extracting and canceling out 21.

$$\frac{1}{4}=-\frac{1}{2}$$

Least common multiple of 4 and 2 is 4. Convert \frac{1}{4} and -\frac{1}{2} to fractions with denominator 4.

$$\frac{1}{4}=-\frac{2}{4}$$

Compare \frac{1}{4} and -\frac{2}{4}.

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