Solve (7x^3y/3)div(-6x^3y^3/4)=

$$( \frac { 7 x ^ { 3 } y } { 3 } ) \div ( - \frac { 6 x ^ { 3 } y ^ { 3 } } { 4 } ) =$$

Evaluate

$$-\frac{14}{9y^{2}}$$

Short Solution Steps

Divide 6x^{3}y^{3} by 4 to get \frac{3}{2}x^{3}y^{3}.

$$\frac{\frac{7x^{3}y}{3}}{-\frac{3}{2}x^{3}y^{3}}$$

Express \frac{\frac{7x^{3}y}{3}}{-\frac{3}{2}x^{3}y^{3}} as a single fraction.

$$\frac{7x^{3}y}{3\left(-\frac{3}{2}x^{3}y^{3}\right)}$$

Factor the expressions that are not already factored.

$$\frac{7yx^{3}}{-\frac{3}{2}\times 3x^{3}y^{3}}$$

Cancel out yx^{3} in both numerator and denominator.

$$\frac{7}{-\frac{3}{2}\times 3y^{2}}$$

Expand the expression.

$$\frac{7}{-\frac{9}{2}y^{2}}$$

Differentiate w.r.t. y

$$\frac{28}{9y^{3}}$$