Solve (1/2x-1/4y)(1/2x+1/4y)+(frac{3}{4}x-y)(frac{3}{4}x+y)

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

Evaluate

$$\frac{13x^{2}-17y^{2}}{16}$$

Solution Steps

Consider \left(\frac{1}{2}x-\frac{1}{4}y\right)\left(\frac{1}{2}x+\frac{1}{4}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}.

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

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

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

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

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

Expand \left(\frac{1}{4}y\right)^{2}.

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

Calculate \frac{1}{4} to the power of 2 and get \frac{1}{16}.

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

Consider \left(\frac{3}{4}x-y\right)\left(\frac{3}{4}x+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}.

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

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

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

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

$$\frac{1}{4}x^{2}-\frac{1}{16}y^{2}+\frac{9}{16}x^{2}-y^{2}$$

Combine \frac{1}{4}x^{2} and \frac{9}{16}x^{2} to get \frac{13}{16}x^{2}.

$$\frac{13}{16}x^{2}-\frac{1}{16}y^{2}-y^{2}$$

Combine -\frac{1}{16}y^{2} and -y^{2} to get -\frac{17}{16}y^{2}.

$$\frac{13}{16}x^{2}-\frac{17}{16}y^{2}$$

Expand

$$\frac{13x^{2}-17y^{2}}{16}$$

Solution Steps

Consider \left(\frac{1}{2}x-\frac{1}{4}y\right)\left(\frac{1}{2}x+\frac{1}{4}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}.

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

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

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

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

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

Expand \left(\frac{1}{4}y\right)^{2}.

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

Calculate \frac{1}{4} to the power of 2 and get \frac{1}{16}.

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

Consider \left(\frac{3}{4}x-y\right)\left(\frac{3}{4}x+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}.

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

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

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

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

$$\frac{1}{4}x^{2}-\frac{1}{16}y^{2}+\frac{9}{16}x^{2}-y^{2}$$

Combine \frac{1}{4}x^{2} and \frac{9}{16}x^{2} to get \frac{13}{16}x^{2}.

$$\frac{13}{16}x^{2}-\frac{1}{16}y^{2}-y^{2}$$

Combine -\frac{1}{16}y^{2} and -y^{2} to get -\frac{17}{16}y^{2}.

$$\frac{13}{16}x^{2}-\frac{17}{16}y^{2}$$