Solve {[(3/2x-9/4x)^{2}*(3y^2-11/3y^2)^2]:[(1/2x-frac{3}{4}x)*(-2y)^3]-frac{3}{8}xy}:(frac{3}{2}x+frac{1}{4}x)

$$\{ [ ( \frac { 3 } { 2 } x - \frac { 9 } { 4 } x ) ^ { 2 } \cdot ( 3 y ^ { 2 } - \frac { 11 } { 3 } y ^ { 2 } ) ^ { 2 } ] : [ ( \frac { 1 } { 2 } x - \frac { 3 } { 4 } x ) \cdot ( - 2 y ) ^ { 3 } ] - \frac { 3 } { 8 } x y \} : ( \frac { 3 } { 2 } x + \frac { 1 } { 4 } x )$$

Evaluate

$$-\frac{y}{7}$$

Solution Steps

Combine \frac{3}{2}x and -\frac{9}{4}x to get -\frac{3}{4}x.

$$\frac{\frac{\left(-\frac{3}{4}x\right)^{2}\left(3y^{2}-\frac{11}{3}y^{2}\right)^{2}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Expand \left(-\frac{3}{4}x\right)^{2}.

$$\frac{\frac{\left(-\frac{3}{4}\right)^{2}x^{2}\left(3y^{2}-\frac{11}{3}y^{2}\right)^{2}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Calculate -\frac{3}{4} to the power of 2 and get \frac{9}{16}.

$$\frac{\frac{\frac{9}{16}x^{2}\left(3y^{2}-\frac{11}{3}y^{2}\right)^{2}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Combine 3y^{2} and -\frac{11}{3}y^{2} to get -\frac{2}{3}y^{2}.

$$\frac{\frac{\frac{9}{16}x^{2}\left(-\frac{2}{3}y^{2}\right)^{2}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Expand \left(-\frac{2}{3}y^{2}\right)^{2}.

$$\frac{\frac{\frac{9}{16}x^{2}\left(-\frac{2}{3}\right)^{2}\left(y^{2}\right)^{2}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\frac{\frac{\frac{9}{16}x^{2}\left(-\frac{2}{3}\right)^{2}y^{4}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Calculate -\frac{2}{3} to the power of 2 and get \frac{4}{9}.

$$\frac{\frac{\frac{9}{16}x^{2}\times \frac{4}{9}y^{4}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Multiply \frac{9}{16} and \frac{4}{9} to get \frac{1}{4}.

$$\frac{\frac{\frac{1}{4}x^{2}y^{4}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Combine \frac{1}{2}x and -\frac{3}{4}x to get -\frac{1}{4}x.

$$\frac{\frac{\frac{1}{4}x^{2}y^{4}}{-\frac{1}{4}x\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Expand \left(-2y\right)^{3}.

$$\frac{\frac{\frac{1}{4}x^{2}y^{4}}{-\frac{1}{4}x\left(-2\right)^{3}y^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Calculate -2 to the power of 3 and get -8.

$$\frac{\frac{\frac{1}{4}x^{2}y^{4}}{-\frac{1}{4}x\left(-8\right)y^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Multiply -\frac{1}{4} and -8 to get 2.

$$\frac{\frac{\frac{1}{4}x^{2}y^{4}}{2xy^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

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

$$\frac{\frac{\frac{1}{4}xy}{2}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Divide \frac{1}{4}xy by 2 to get \frac{1}{8}xy.

$$\frac{\frac{1}{8}xy-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Combine \frac{1}{8}xy and -\frac{3}{8}xy to get -\frac{1}{4}xy.

$$\frac{-\frac{1}{4}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Combine \frac{3}{2}x and \frac{1}{4}x to get \frac{7}{4}x.

$$\frac{-\frac{1}{4}xy}{\frac{7}{4}x}$$

Cancel out x in both numerator and denominator.

$$\frac{-\frac{1}{4}y}{\frac{7}{4}}$$

Divide -\frac{1}{4}y by \frac{7}{4} by multiplying -\frac{1}{4}y by the reciprocal of \frac{7}{4}.

$$\frac{-\frac{1}{4}y\times 4}{7}$$

Multiply -\frac{1}{4} and 4 to get -1.

$$\frac{-y}{7}$$

Expand

$$-\frac{y}{7}$$

Solution Steps

Combine \frac{3}{2}x and -\frac{9}{4}x to get -\frac{3}{4}x.

$$\frac{\frac{\left(-\frac{3}{4}x\right)^{2}\left(3y^{2}-\frac{11}{3}y^{2}\right)^{2}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Expand \left(-\frac{3}{4}x\right)^{2}.

$$\frac{\frac{\left(-\frac{3}{4}\right)^{2}x^{2}\left(3y^{2}-\frac{11}{3}y^{2}\right)^{2}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Calculate -\frac{3}{4} to the power of 2 and get \frac{9}{16}.

$$\frac{\frac{\frac{9}{16}x^{2}\left(3y^{2}-\frac{11}{3}y^{2}\right)^{2}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Combine 3y^{2} and -\frac{11}{3}y^{2} to get -\frac{2}{3}y^{2}.

$$\frac{\frac{\frac{9}{16}x^{2}\left(-\frac{2}{3}y^{2}\right)^{2}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Expand \left(-\frac{2}{3}y^{2}\right)^{2}.

$$\frac{\frac{\frac{9}{16}x^{2}\left(-\frac{2}{3}\right)^{2}\left(y^{2}\right)^{2}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\frac{\frac{\frac{9}{16}x^{2}\left(-\frac{2}{3}\right)^{2}y^{4}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Calculate -\frac{2}{3} to the power of 2 and get \frac{4}{9}.

$$\frac{\frac{\frac{9}{16}x^{2}\times \frac{4}{9}y^{4}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Multiply \frac{9}{16} and \frac{4}{9} to get \frac{1}{4}.

$$\frac{\frac{\frac{1}{4}x^{2}y^{4}}{\left(\frac{1}{2}x-\frac{3}{4}x\right)\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Combine \frac{1}{2}x and -\frac{3}{4}x to get -\frac{1}{4}x.

$$\frac{\frac{\frac{1}{4}x^{2}y^{4}}{-\frac{1}{4}x\left(-2y\right)^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Expand \left(-2y\right)^{3}.

$$\frac{\frac{\frac{1}{4}x^{2}y^{4}}{-\frac{1}{4}x\left(-2\right)^{3}y^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Calculate -2 to the power of 3 and get -8.

$$\frac{\frac{\frac{1}{4}x^{2}y^{4}}{-\frac{1}{4}x\left(-8\right)y^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Multiply -\frac{1}{4} and -8 to get 2.

$$\frac{\frac{\frac{1}{4}x^{2}y^{4}}{2xy^{3}}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

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

$$\frac{\frac{\frac{1}{4}xy}{2}-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Divide \frac{1}{4}xy by 2 to get \frac{1}{8}xy.

$$\frac{\frac{1}{8}xy-\frac{3}{8}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Combine \frac{1}{8}xy and -\frac{3}{8}xy to get -\frac{1}{4}xy.

$$\frac{-\frac{1}{4}xy}{\frac{3}{2}x+\frac{1}{4}x}$$

Combine \frac{3}{2}x and \frac{1}{4}x to get \frac{7}{4}x.

$$\frac{-\frac{1}{4}xy}{\frac{7}{4}x}$$

Cancel out x in both numerator and denominator.

$$\frac{-\frac{1}{4}y}{\frac{7}{4}}$$

Divide -\frac{1}{4}y by \frac{7}{4} by multiplying -\frac{1}{4}y by the reciprocal of \frac{7}{4}.

$$\frac{-\frac{1}{4}y\times 4}{7}$$

Multiply -\frac{1}{4} and 4 to get -1.

$$\frac{-y}{7}$$