Solve (frac{15sqrt{23}}{16})^2+(93/16-x)^2=(x+y)^2

$$( \frac { 15 \sqrt { 23 } } { 16 } ) ^ { 2 } + ( \frac { 93 } { 16 } - x ) ^ { 2 } = ( x + y ) ^ { 2 }$$

Solve for x

$$x=\frac{8\left(54-y^{2}\right)}{16y+93}$$

$$y\neq -\frac{93}{16}$$

Steps for Solving Linear Equation

To raise \frac{15\sqrt{23}}{16} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(15\sqrt{23}\right)^{2}}{16^{2}}+\left(\frac{93}{16}-x\right)^{2}=\left(x+y\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{93}{16}-x\right)^{2}.

$$\frac{\left(15\sqrt{23}\right)^{2}}{16^{2}}+\frac{8649}{256}-\frac{93}{8}x+x^{2}=\left(x+y\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 16^{2}.

$$\frac{\left(15\sqrt{23}\right)^{2}}{256}+\frac{8649}{256}-\frac{93}{8}x+x^{2}=\left(x+y\right)^{2}$$

Since \frac{\left(15\sqrt{23}\right)^{2}}{256} and \frac{8649}{256} have the same denominator, add them by adding their numerators.

$$\frac{\left(15\sqrt{23}\right)^{2}+8649}{256}-\frac{93}{8}x+x^{2}=\left(x+y\right)^{2}$$

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

$$\frac{\left(15\sqrt{23}\right)^{2}+8649}{256}-\frac{93}{8}x+x^{2}=x^{2}+2xy+y^{2}$$

Expand \left(15\sqrt{23}\right)^{2}.

$$\frac{15^{2}\left(\sqrt{23}\right)^{2}+8649}{256}-\frac{93}{8}x+x^{2}=x^{2}+2xy+y^{2}$$

Calculate 15 to the power of 2 and get 225.

$$\frac{225\left(\sqrt{23}\right)^{2}+8649}{256}-\frac{93}{8}x+x^{2}=x^{2}+2xy+y^{2}$$

The square of \sqrt{23} is 23.

$$\frac{225\times 23+8649}{256}-\frac{93}{8}x+x^{2}=x^{2}+2xy+y^{2}$$

Multiply 225 and 23 to get 5175.

$$\frac{5175+8649}{256}-\frac{93}{8}x+x^{2}=x^{2}+2xy+y^{2}$$

Add 5175 and 8649 to get 13824.

$$\frac{13824}{256}-\frac{93}{8}x+x^{2}=x^{2}+2xy+y^{2}$$

Divide 13824 by 256 to get 54.

$$54-\frac{93}{8}x+x^{2}=x^{2}+2xy+y^{2}$$

Subtract x^{2} from both sides.

$$54-\frac{93}{8}x+x^{2}-x^{2}=2xy+y^{2}$$

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

$$54-\frac{93}{8}x=2xy+y^{2}$$

Subtract 2xy from both sides.

$$54-\frac{93}{8}x-2xy=y^{2}$$

Subtract 54 from both sides.

$$-\frac{93}{8}x-2xy=y^{2}-54$$

Combine all terms containing x.

$$\left(-\frac{93}{8}-2y\right)x=y^{2}-54$$

The equation is in standard form.

$$\left(-2y-\frac{93}{8}\right)x=y^{2}-54$$

Divide both sides by -2y-\frac{93}{8}.

$$\frac{\left(-2y-\frac{93}{8}\right)x}{-2y-\frac{93}{8}}=\frac{y^{2}-54}{-2y-\frac{93}{8}}$$

Dividing by -2y-\frac{93}{8} undoes the multiplication by -2y-\frac{93}{8}.

$$x=\frac{y^{2}-54}{-2y-\frac{93}{8}}$$

Divide y^{2}-54 by -2y-\frac{93}{8}.

$$x=-\frac{8\left(y^{2}-54\right)}{16y+93}$$

Solve for y

$$y=\frac{\sqrt{16x^{2}-186x+864}}{4}-x$$

$$y=-\frac{\sqrt{16x^{2}-186x+864}}{4}-x$$