Solve 33/8x^2+4(5/4x-2y)(5/4x+2y)-[5xy+(5/4x-2y)^2-24y^2]-[(frac{3}{4}x)^2+4y(y-1)]

$$\frac { 33 } { 8 } x ^ { 2 } + 4 ( \frac { 5 } { 4 } x - 2 y ) ( \frac { 5 } { 4 } x + 2 y ) - [ 5 x y + ( \frac { 5 } { 4 } x - 2 y ) ^ { 2 } - 24 y ^ { 2 } ] - [ ( \frac { 3 } { 4 } x ) ^ { 2 } + 4 y ( y - 1 ) ]$$

Evaluate

$$\frac{33x^{2}}{4}+4y$$

Solution Steps

Use the distributive property to multiply 4 by \frac{5}{4}x-2y.

$$\frac{33}{8}x^{2}+\left(5x-8y\right)\left(\frac{5}{4}x+2y\right)-\left(5xy+\left(\frac{5}{4}x-2y\right)^{2}-24y^{2}\right)-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Use the distributive property to multiply 5x-8y by \frac{5}{4}x+2y and combine like terms.

$$\frac{33}{8}x^{2}+\frac{25}{4}x^{2}-16y^{2}-\left(5xy+\left(\frac{5}{4}x-2y\right)^{2}-24y^{2}\right)-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Combine \frac{33}{8}x^{2} and \frac{25}{4}x^{2} to get \frac{83}{8}x^{2}.

$$\frac{83}{8}x^{2}-16y^{2}-\left(5xy+\left(\frac{5}{4}x-2y\right)^{2}-24y^{2}\right)-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{5}{4}x-2y\right)^{2}.

$$\frac{83}{8}x^{2}-16y^{2}-\left(5xy+\frac{25}{16}x^{2}-5xy+4y^{2}-24y^{2}\right)-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Combine 5xy and -5xy to get 0.

$$\frac{83}{8}x^{2}-16y^{2}-\left(\frac{25}{16}x^{2}+4y^{2}-24y^{2}\right)-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Combine 4y^{2} and -24y^{2} to get -20y^{2}.

$$\frac{83}{8}x^{2}-16y^{2}-\left(\frac{25}{16}x^{2}-20y^{2}\right)-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

To find the opposite of \frac{25}{16}x^{2}-20y^{2}, find the opposite of each term.

$$\frac{83}{8}x^{2}-16y^{2}-\frac{25}{16}x^{2}+20y^{2}-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Combine \frac{83}{8}x^{2} and -\frac{25}{16}x^{2} to get \frac{141}{16}x^{2}.

$$\frac{141}{16}x^{2}-16y^{2}+20y^{2}-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Combine -16y^{2} and 20y^{2} to get 4y^{2}.

$$\frac{141}{16}x^{2}+4y^{2}-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

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

$$\frac{141}{16}x^{2}+4y^{2}-\left(\left(\frac{3}{4}\right)^{2}x^{2}+4y\left(y-1\right)\right)$$

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

$$\frac{141}{16}x^{2}+4y^{2}-\left(\frac{9}{16}x^{2}+4y\left(y-1\right)\right)$$

Use the distributive property to multiply 4y by y-1.

$$\frac{141}{16}x^{2}+4y^{2}-\left(\frac{9}{16}x^{2}+4y^{2}-4y\right)$$

To find the opposite of \frac{9}{16}x^{2}+4y^{2}-4y, find the opposite of each term.

$$\frac{141}{16}x^{2}+4y^{2}-\frac{9}{16}x^{2}-4y^{2}+4y$$

Combine \frac{141}{16}x^{2} and -\frac{9}{16}x^{2} to get \frac{33}{4}x^{2}.

$$\frac{33}{4}x^{2}+4y^{2}-4y^{2}+4y$$

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

$$\frac{33}{4}x^{2}+4y$$

Expand

$$\frac{33x^{2}}{4}+4y$$

Solution Steps

Use the distributive property to multiply 4 by \frac{5}{4}x-2y.

$$\frac{33}{8}x^{2}+\left(5x-8y\right)\left(\frac{5}{4}x+2y\right)-\left(5xy+\left(\frac{5}{4}x-2y\right)^{2}-24y^{2}\right)-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Use the distributive property to multiply 5x-8y by \frac{5}{4}x+2y and combine like terms.

$$\frac{33}{8}x^{2}+\frac{25}{4}x^{2}-16y^{2}-\left(5xy+\left(\frac{5}{4}x-2y\right)^{2}-24y^{2}\right)-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Combine \frac{33}{8}x^{2} and \frac{25}{4}x^{2} to get \frac{83}{8}x^{2}.

$$\frac{83}{8}x^{2}-16y^{2}-\left(5xy+\left(\frac{5}{4}x-2y\right)^{2}-24y^{2}\right)-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{5}{4}x-2y\right)^{2}.

$$\frac{83}{8}x^{2}-16y^{2}-\left(5xy+\frac{25}{16}x^{2}-5xy+4y^{2}-24y^{2}\right)-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Combine 5xy and -5xy to get 0.

$$\frac{83}{8}x^{2}-16y^{2}-\left(\frac{25}{16}x^{2}+4y^{2}-24y^{2}\right)-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Combine 4y^{2} and -24y^{2} to get -20y^{2}.

$$\frac{83}{8}x^{2}-16y^{2}-\left(\frac{25}{16}x^{2}-20y^{2}\right)-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

To find the opposite of \frac{25}{16}x^{2}-20y^{2}, find the opposite of each term.

$$\frac{83}{8}x^{2}-16y^{2}-\frac{25}{16}x^{2}+20y^{2}-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Combine \frac{83}{8}x^{2} and -\frac{25}{16}x^{2} to get \frac{141}{16}x^{2}.

$$\frac{141}{16}x^{2}-16y^{2}+20y^{2}-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

Combine -16y^{2} and 20y^{2} to get 4y^{2}.

$$\frac{141}{16}x^{2}+4y^{2}-\left(\left(\frac{3}{4}x\right)^{2}+4y\left(y-1\right)\right)$$

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

$$\frac{141}{16}x^{2}+4y^{2}-\left(\left(\frac{3}{4}\right)^{2}x^{2}+4y\left(y-1\right)\right)$$

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

$$\frac{141}{16}x^{2}+4y^{2}-\left(\frac{9}{16}x^{2}+4y\left(y-1\right)\right)$$

Use the distributive property to multiply 4y by y-1.

$$\frac{141}{16}x^{2}+4y^{2}-\left(\frac{9}{16}x^{2}+4y^{2}-4y\right)$$

To find the opposite of \frac{9}{16}x^{2}+4y^{2}-4y, find the opposite of each term.

$$\frac{141}{16}x^{2}+4y^{2}-\frac{9}{16}x^{2}-4y^{2}+4y$$

Combine \frac{141}{16}x^{2} and -\frac{9}{16}x^{2} to get \frac{33}{4}x^{2}.

$$\frac{33}{4}x^{2}+4y^{2}-4y^{2}+4y$$

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

$$\frac{33}{4}x^{2}+4y$$