Solve -y^2+y^2-10x-8y+32=0

$$- y ^ { 2 } + y ^ { 2 } - 10 x - 8 y + 32 = 0$$

Solve for x

$$x=\frac{16-4y}{5}$$

Steps for Solving Linear Equation

Add y^{2} to both sides. Anything plus zero gives itself.

$$y^{2}-10x-8y+32=y^{2}$$

Subtract y^{2} from both sides.

$$-10x-8y+32=y^{2}-y^{2}$$

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

$$-10x-8y+32=0$$

Add 8y to both sides. Anything plus zero gives itself.

$$-10x+32=8y$$

Subtract 32 from both sides.

$$-10x=8y-32$$

Divide both sides by -10.

$$\frac{-10x}{-10}=\frac{8y-32}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x=\frac{8y-32}{-10}$$

Divide -32+8y by -10.

$$x=\frac{16-4y}{5}$$

Solve for y

$$y=-\frac{5x}{4}+4$$

Steps for Solving Linear Equation

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

$$-10x-8y+32=0$$

Add 10x to both sides. Anything plus zero gives itself.

$$-8y+32=10x$$

Subtract 32 from both sides.

$$-8y=10x-32$$

Divide both sides by -8.

$$\frac{-8y}{-8}=\frac{10x-32}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$y=\frac{10x-32}{-8}$$

Divide 10x-32 by -8.

$$y=-\frac{5x}{4}+4$$