Solve 31/5:(3/7-14/5)-4/3*2

$$3 \frac { 1 } { 5 } : ( \frac { 3 } { 7 } - 1 \frac { 4 } { 5 } ) - \frac { 4 } { 3 } \cdot 2$$

Evaluate

$$-5$$

Solution Steps

Multiply 3 and 5 to get 15.

$$\frac{\frac{15+1}{5}}{\frac{3}{7}-\frac{1\times 5+4}{5}}-\frac{4}{3}\times 2$$

Add 15 and 1 to get 16.

$$\frac{\frac{16}{5}}{\frac{3}{7}-\frac{1\times 5+4}{5}}-\frac{4}{3}\times 2$$

Multiply 1 and 5 to get 5.

$$\frac{\frac{16}{5}}{\frac{3}{7}-\frac{5+4}{5}}-\frac{4}{3}\times 2$$

Add 5 and 4 to get 9.

$$\frac{\frac{16}{5}}{\frac{3}{7}-\frac{9}{5}}-\frac{4}{3}\times 2$$

Least common multiple of 7 and 5 is 35. Convert \frac{3}{7} and \frac{9}{5} to fractions with denominator 35.

$$\frac{\frac{16}{5}}{\frac{15}{35}-\frac{63}{35}}-\frac{4}{3}\times 2$$

Since \frac{15}{35} and \frac{63}{35} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{16}{5}}{\frac{15-63}{35}}-\frac{4}{3}\times 2$$

Subtract 63 from 15 to get -48.

$$\frac{\frac{16}{5}}{-\frac{48}{35}}-\frac{4}{3}\times 2$$

Divide \frac{16}{5} by -\frac{48}{35} by multiplying \frac{16}{5} by the reciprocal of -\frac{48}{35}.

$$\frac{16}{5}\left(-\frac{35}{48}\right)-\frac{4}{3}\times 2$$

Multiply \frac{16}{5} times -\frac{35}{48} by multiplying numerator times numerator and denominator times denominator.

$$\frac{16\left(-35\right)}{5\times 48}-\frac{4}{3}\times 2$$

Do the multiplications in the fraction \frac{16\left(-35\right)}{5\times 48}.

$$\frac{-560}{240}-\frac{4}{3}\times 2$$

Reduce the fraction \frac{-560}{240} to lowest terms by extracting and canceling out 80.

$$-\frac{7}{3}-\frac{4}{3}\times 2$$

Express \frac{4}{3}\times 2 as a single fraction.

$$-\frac{7}{3}-\frac{4\times 2}{3}$$

Multiply 4 and 2 to get 8.

$$-\frac{7}{3}-\frac{8}{3}$$

Since -\frac{7}{3} and \frac{8}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-7-8}{3}$$

Subtract 8 from -7 to get -15.

$$\frac{-15}{3}$$

Divide -15 by 3 to get -5.

$$-5$$

Factor

$$-5$$