Solve (2x+1/3y)(x-3y)-(2x+y)(1/2x-y)=

$$( 2 x + \frac { 1 } { 3 } y ) ( x - 3 y ) - ( 2 x + y ) ( \frac { 1 } { 2 } x - y ) =$$

Evaluate

$$-\frac{25xy}{6}+x^{2}$$

Solution Steps

Apply the distributive property by multiplying each term of 2x+\frac{1}{3}y by each term of x-3y.

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

Multiply y and y to get y^{2}.

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

Combine -6xy and \frac{1}{3}yx to get -\frac{17}{3}xy.

$$2x^{2}-\frac{17}{3}xy+\frac{1}{3}y^{2}\left(-3\right)-\left(2x+y\right)\left(\frac{1}{2}x-y\right)$$

Multiply \frac{1}{3} and -3 to get \frac{-3}{3}.

$$2x^{2}-\frac{17}{3}xy+\frac{-3}{3}y^{2}-\left(2x+y\right)\left(\frac{1}{2}x-y\right)$$

Divide -3 by 3 to get -1.

$$2x^{2}-\frac{17}{3}xy-y^{2}-\left(2x+y\right)\left(\frac{1}{2}x-y\right)$$

Apply the distributive property by multiplying each term of 2x+y by each term of \frac{1}{2}x-y.

$$2x^{2}-\frac{17}{3}xy-y^{2}-\left(2x\times \frac{1}{2}x-2xy+y\times \frac{1}{2}x-y^{2}\right)$$

Multiply x and x to get x^{2}.

$$2x^{2}-\frac{17}{3}xy-y^{2}-\left(2x^{2}\times \frac{1}{2}-2xy+y\times \frac{1}{2}x-y^{2}\right)$$

Cancel out 2 and 2.

$$2x^{2}-\frac{17}{3}xy-y^{2}-\left(x^{2}-2xy+y\times \frac{1}{2}x-y^{2}\right)$$

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

$$2x^{2}-\frac{17}{3}xy-y^{2}-\left(x^{2}-\frac{3}{2}xy-y^{2}\right)$$

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

$$2x^{2}-\frac{17}{3}xy-y^{2}-x^{2}-\left(-\frac{3}{2}xy\right)-\left(-y^{2}\right)$$

The opposite of -\frac{3}{2}xy is \frac{3}{2}xy.

$$2x^{2}-\frac{17}{3}xy-y^{2}-x^{2}+\frac{3}{2}xy-\left(-y^{2}\right)$$

The opposite of -y^{2} is y^{2}.

$$2x^{2}-\frac{17}{3}xy-y^{2}-x^{2}+\frac{3}{2}xy+y^{2}$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$x^{2}-\frac{17}{3}xy-y^{2}+\frac{3}{2}xy+y^{2}$$

Combine -\frac{17}{3}xy and \frac{3}{2}xy to get -\frac{25}{6}xy.

$$x^{2}-\frac{25}{6}xy-y^{2}+y^{2}$$

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

$$x^{2}-\frac{25}{6}xy$$

Expand

$$-\frac{25xy}{6}+x^{2}$$

Solution Steps

Apply the distributive property by multiplying each term of 2x+\frac{1}{3}y by each term of x-3y.

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

Multiply y and y to get y^{2}.

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

Combine -6xy and \frac{1}{3}yx to get -\frac{17}{3}xy.

$$2x^{2}-\frac{17}{3}xy+\frac{1}{3}y^{2}\left(-3\right)-\left(2x+y\right)\left(\frac{1}{2}x-y\right)$$

Multiply \frac{1}{3} and -3 to get \frac{-3}{3}.

$$2x^{2}-\frac{17}{3}xy+\frac{-3}{3}y^{2}-\left(2x+y\right)\left(\frac{1}{2}x-y\right)$$

Divide -3 by 3 to get -1.

$$2x^{2}-\frac{17}{3}xy-y^{2}-\left(2x+y\right)\left(\frac{1}{2}x-y\right)$$

Apply the distributive property by multiplying each term of 2x+y by each term of \frac{1}{2}x-y.

$$2x^{2}-\frac{17}{3}xy-y^{2}-\left(2x\times \frac{1}{2}x-2xy+y\times \frac{1}{2}x-y^{2}\right)$$

Multiply x and x to get x^{2}.

$$2x^{2}-\frac{17}{3}xy-y^{2}-\left(2x^{2}\times \frac{1}{2}-2xy+y\times \frac{1}{2}x-y^{2}\right)$$

Cancel out 2 and 2.

$$2x^{2}-\frac{17}{3}xy-y^{2}-\left(x^{2}-2xy+y\times \frac{1}{2}x-y^{2}\right)$$

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

$$2x^{2}-\frac{17}{3}xy-y^{2}-\left(x^{2}-\frac{3}{2}xy-y^{2}\right)$$

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

$$2x^{2}-\frac{17}{3}xy-y^{2}-x^{2}-\left(-\frac{3}{2}xy\right)-\left(-y^{2}\right)$$

The opposite of -\frac{3}{2}xy is \frac{3}{2}xy.

$$2x^{2}-\frac{17}{3}xy-y^{2}-x^{2}+\frac{3}{2}xy-\left(-y^{2}\right)$$

The opposite of -y^{2} is y^{2}.

$$2x^{2}-\frac{17}{3}xy-y^{2}-x^{2}+\frac{3}{2}xy+y^{2}$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$x^{2}-\frac{17}{3}xy-y^{2}+\frac{3}{2}xy+y^{2}$$

Combine -\frac{17}{3}xy and \frac{3}{2}xy to get -\frac{25}{6}xy.

$$x^{2}-\frac{25}{6}xy-y^{2}+y^{2}$$

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

$$x^{2}-\frac{25}{6}xy$$