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

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

Evaluate

$$y^{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{3}x-\frac{1}{2}y\right)^{2}.

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

Use the distributive property to multiply 4 by \frac{1}{9}x^{2}-\frac{1}{3}xy+\frac{1}{4}y^{2}.

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

Combine \frac{4}{3}xy and -\frac{4}{3}xy to get 0.

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

Consider \left(\frac{2}{3}x-\frac{1}{2}y\right)\left(\frac{2}{3}x+\frac{1}{2}y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

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

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

Expand \left(\frac{1}{2}y\right)^{2}.

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

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

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

To find the opposite of \frac{4}{9}x^{2}-\frac{1}{4}y^{2}, find the opposite of each term.

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

Combine \frac{4}{9}x^{2} and -\frac{4}{9}x^{2} to get 0.

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

Combine y^{2} and \frac{1}{4}y^{2} to get \frac{5}{4}y^{2}.

$$2\times \frac{5}{4}y^{2}-\frac{3}{2}y^{2}$$

Multiply 2 and \frac{5}{4} to get \frac{5}{2}.

$$\frac{5}{2}y^{2}-\frac{3}{2}y^{2}$$

Combine \frac{5}{2}y^{2} and -\frac{3}{2}y^{2} to get y^{2}.

$$y^{2}$$

Expand

$$y^{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{3}x-\frac{1}{2}y\right)^{2}.

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

Use the distributive property to multiply 4 by \frac{1}{9}x^{2}-\frac{1}{3}xy+\frac{1}{4}y^{2}.

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

Combine \frac{4}{3}xy and -\frac{4}{3}xy to get 0.

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

Consider \left(\frac{2}{3}x-\frac{1}{2}y\right)\left(\frac{2}{3}x+\frac{1}{2}y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

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

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

Expand \left(\frac{1}{2}y\right)^{2}.

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

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

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

To find the opposite of \frac{4}{9}x^{2}-\frac{1}{4}y^{2}, find the opposite of each term.

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

Combine \frac{4}{9}x^{2} and -\frac{4}{9}x^{2} to get 0.

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

Combine y^{2} and \frac{1}{4}y^{2} to get \frac{5}{4}y^{2}.

$$2\times \frac{5}{4}y^{2}-\frac{3}{2}y^{2}$$

Multiply 2 and \frac{5}{4} to get \frac{5}{2}.

$$\frac{5}{2}y^{2}-\frac{3}{2}y^{2}$$

Combine \frac{5}{2}y^{2} and -\frac{3}{2}y^{2} to get y^{2}.

$$y^{2}$$