Solve (5/12x^{2}-1/3xy)-{2y^2-[(1/3x^2+3/2xy)-(2xy-2y^2+frac{1}{4}x^2)-xy]}-frac{1}{6}xy

$$( \frac { 5 } { 12 } x ^ { 2 } - \frac { 1 } { 3 } x y ) - \{ 2 y ^ { 2 } - [ ( \frac { 1 } { 3 } x ^ { 2 } + \frac { 3 } { 2 } x y ) - ( 2 x y - 2 y ^ { 2 } + \frac { 1 } { 4 } x ^ { 2 } ) - x y ] \} - \frac { 1 } { 6 } x y$$

Evaluate

$$\frac{x\left(x-4y\right)}{2}$$

Solution Steps

To find the opposite of 2xy-2y^{2}+\frac{1}{4}x^{2}, find the opposite of each term.

$$\frac{5}{12}x^{2}-\frac{1}{3}xy-\left(2y^{2}-\left(\frac{1}{3}x^{2}+\frac{3}{2}xy-2xy+2y^{2}-\frac{1}{4}x^{2}-xy\right)\right)-\frac{1}{6}xy$$

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

$$\frac{5}{12}x^{2}-\frac{1}{3}xy-\left(2y^{2}-\left(\frac{1}{3}x^{2}-\frac{1}{2}xy+2y^{2}-\frac{1}{4}x^{2}-xy\right)\right)-\frac{1}{6}xy$$

Combine \frac{1}{3}x^{2} and -\frac{1}{4}x^{2} to get \frac{1}{12}x^{2}.

$$\frac{5}{12}x^{2}-\frac{1}{3}xy-\left(2y^{2}-\left(\frac{1}{12}x^{2}-\frac{1}{2}xy+2y^{2}-xy\right)\right)-\frac{1}{6}xy$$

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

$$\frac{5}{12}x^{2}-\frac{1}{3}xy-\left(2y^{2}-\left(\frac{1}{12}x^{2}-\frac{3}{2}xy+2y^{2}\right)\right)-\frac{1}{6}xy$$

To find the opposite of \frac{1}{12}x^{2}-\frac{3}{2}xy+2y^{2}, find the opposite of each term.

$$\frac{5}{12}x^{2}-\frac{1}{3}xy-\left(2y^{2}-\frac{1}{12}x^{2}+\frac{3}{2}xy-2y^{2}\right)-\frac{1}{6}xy$$

Combine 2y^{2} and -2y^{2} to get 0.

$$\frac{5}{12}x^{2}-\frac{1}{3}xy-\left(-\frac{1}{12}x^{2}+\frac{3}{2}xy\right)-\frac{1}{6}xy$$

To find the opposite of -\frac{1}{12}x^{2}+\frac{3}{2}xy, find the opposite of each term.

$$\frac{5}{12}x^{2}-\frac{1}{3}xy+\frac{1}{12}x^{2}-\frac{3}{2}xy-\frac{1}{6}xy$$

Combine \frac{5}{12}x^{2} and \frac{1}{12}x^{2} to get \frac{1}{2}x^{2}.

$$\frac{1}{2}x^{2}-\frac{1}{3}xy-\frac{3}{2}xy-\frac{1}{6}xy$$

Combine -\frac{1}{3}xy and -\frac{3}{2}xy to get -\frac{11}{6}xy.

$$\frac{1}{2}x^{2}-\frac{11}{6}xy-\frac{1}{6}xy$$

Combine -\frac{11}{6}xy and -\frac{1}{6}xy to get -2xy.

$$\frac{1}{2}x^{2}-2xy$$

Expand

$$-2xy+\frac{x^{2}}{2}$$

Solution Steps

To find the opposite of 2xy-2y^{2}+\frac{1}{4}x^{2}, find the opposite of each term.

$$\frac{5}{12}x^{2}-\frac{1}{3}xy-\left(2y^{2}-\left(\frac{1}{3}x^{2}+\frac{3}{2}xy-2xy+2y^{2}-\frac{1}{4}x^{2}-xy\right)\right)-\frac{1}{6}xy$$

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

$$\frac{5}{12}x^{2}-\frac{1}{3}xy-\left(2y^{2}-\left(\frac{1}{3}x^{2}-\frac{1}{2}xy+2y^{2}-\frac{1}{4}x^{2}-xy\right)\right)-\frac{1}{6}xy$$

Combine \frac{1}{3}x^{2} and -\frac{1}{4}x^{2} to get \frac{1}{12}x^{2}.

$$\frac{5}{12}x^{2}-\frac{1}{3}xy-\left(2y^{2}-\left(\frac{1}{12}x^{2}-\frac{1}{2}xy+2y^{2}-xy\right)\right)-\frac{1}{6}xy$$

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

$$\frac{5}{12}x^{2}-\frac{1}{3}xy-\left(2y^{2}-\left(\frac{1}{12}x^{2}-\frac{3}{2}xy+2y^{2}\right)\right)-\frac{1}{6}xy$$

To find the opposite of \frac{1}{12}x^{2}-\frac{3}{2}xy+2y^{2}, find the opposite of each term.

$$\frac{5}{12}x^{2}-\frac{1}{3}xy-\left(2y^{2}-\frac{1}{12}x^{2}+\frac{3}{2}xy-2y^{2}\right)-\frac{1}{6}xy$$

Combine 2y^{2} and -2y^{2} to get 0.

$$\frac{5}{12}x^{2}-\frac{1}{3}xy-\left(-\frac{1}{12}x^{2}+\frac{3}{2}xy\right)-\frac{1}{6}xy$$

To find the opposite of -\frac{1}{12}x^{2}+\frac{3}{2}xy, find the opposite of each term.

$$\frac{5}{12}x^{2}-\frac{1}{3}xy+\frac{1}{12}x^{2}-\frac{3}{2}xy-\frac{1}{6}xy$$

Combine \frac{5}{12}x^{2} and \frac{1}{12}x^{2} to get \frac{1}{2}x^{2}.

$$\frac{1}{2}x^{2}-\frac{1}{3}xy-\frac{3}{2}xy-\frac{1}{6}xy$$

Combine -\frac{1}{3}xy and -\frac{3}{2}xy to get -\frac{11}{6}xy.

$$\frac{1}{2}x^{2}-\frac{11}{6}xy-\frac{1}{6}xy$$

Combine -\frac{11}{6}xy and -\frac{1}{6}xy to get -2xy.

$$\frac{1}{2}x^{2}-2xy$$