Solve (x-1/2y)(x-1/2y)-1-(x-1/2y-1)(x-1/2y-1)

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

Evaluate

$$2x-y-2$$

Solution Steps

Multiply x-\frac{1}{2}y and x-\frac{1}{2}y to get \left(x-\frac{1}{2}y\right)^{2}.

$$\left(x-\frac{1}{2}y\right)^{2}-1-\left(x-\frac{1}{2}y-1\right)\left(x-\frac{1}{2}y-1\right)$$

Multiply x-\frac{1}{2}y-1 and x-\frac{1}{2}y-1 to get \left(x-\frac{1}{2}y-1\right)^{2}.

$$\left(x-\frac{1}{2}y\right)^{2}-1-\left(x-\frac{1}{2}y-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{1}{2}y\right)^{2}.

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

Square x-\frac{1}{2}y-1.

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

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

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

The opposite of -xy is xy.

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

The opposite of -2x is 2x.

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

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

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

Combine -xy and xy to get 0.

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

Combine \frac{1}{4}y^{2} and -\frac{1}{4}y^{2} to get 0.

$$-1+2x-y-1$$

Subtract 1 from -1 to get -2.

$$-2+2x-y$$

Expand

$$2x-y-2$$

Solution Steps

Multiply x-\frac{1}{2}y and x-\frac{1}{2}y to get \left(x-\frac{1}{2}y\right)^{2}.

$$\left(x-\frac{1}{2}y\right)^{2}-1-\left(x-\frac{1}{2}y-1\right)\left(x-\frac{1}{2}y-1\right)$$

Multiply x-\frac{1}{2}y-1 and x-\frac{1}{2}y-1 to get \left(x-\frac{1}{2}y-1\right)^{2}.

$$\left(x-\frac{1}{2}y\right)^{2}-1-\left(x-\frac{1}{2}y-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{1}{2}y\right)^{2}.

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

Square x-\frac{1}{2}y-1.

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

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

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

The opposite of -xy is xy.

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

The opposite of -2x is 2x.

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

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

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

Combine -xy and xy to get 0.

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

Combine \frac{1}{4}y^{2} and -\frac{1}{4}y^{2} to get 0.

$$-1+2x-y-1$$

Subtract 1 from -1 to get -2.

$$-2+2x-y$$