Solve {[(1/7-7/3)*1/5]-4/3}

$$\{ [ ( \frac { 1 } { 7 } - \frac { 7 } { 3 } ) * \frac { 1 } { 5 } ] - \frac { 4 } { 3 } \}$$

Evaluate

$$-\frac{62}{35}\approx -1.771428571$$

Solution Steps

Least common multiple of 7 and 3 is 21. Convert \frac{1}{7} and \frac{7}{3} to fractions with denominator 21.

$$\left(\frac{3}{21}-\frac{49}{21}\right)\times \frac{1}{5}-\frac{4}{3}$$

Since \frac{3}{21} and \frac{49}{21} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3-49}{21}\times \frac{1}{5}-\frac{4}{3}$$

Subtract 49 from 3 to get -46.

$$-\frac{46}{21}\times \frac{1}{5}-\frac{4}{3}$$

Multiply -\frac{46}{21} times \frac{1}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-46}{21\times 5}-\frac{4}{3}$$

Do the multiplications in the fraction \frac{-46}{21\times 5}.

$$\frac{-46}{105}-\frac{4}{3}$$

Fraction \frac{-46}{105} can be rewritten as -\frac{46}{105} by extracting the negative sign.

$$-\frac{46}{105}-\frac{4}{3}$$

Least common multiple of 105 and 3 is 105. Convert -\frac{46}{105} and \frac{4}{3} to fractions with denominator 105.

$$-\frac{46}{105}-\frac{140}{105}$$

Since -\frac{46}{105} and \frac{140}{105} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-46-140}{105}$$

Subtract 140 from -46 to get -186.

$$\frac{-186}{105}$$

Reduce the fraction \frac{-186}{105} to lowest terms by extracting and canceling out 3.

$$-\frac{62}{35}$$

Factor

$$-\frac{62}{35} = -1\frac{27}{35} = -1.7714285714285714$$