Solve x^2-10x+y^2-12y-39=x^2+2x+y^2+4y-95

$${ x }^{ 2 } -10x+ { y }^{ 2 } -12y-39= { x }^{ 2 } +2x+ { y }^{ 2 } +4y-95$$

Solve for x

$$x=\frac{14-4y}{3}$$

Steps for Solving Linear Equation

Subtract x^{2} from both sides.

$$x^{2}-10x+y^{2}-12y-39-x^{2}=2x+y^{2}+4y-95$$

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

$$-10x+y^{2}-12y-39=2x+y^{2}+4y-95$$

Subtract 2x from both sides.

$$-10x+y^{2}-12y-39-2x=y^{2}+4y-95$$

Combine -10x and -2x to get -12x.

$$-12x+y^{2}-12y-39=y^{2}+4y-95$$

Subtract y^{2} from both sides.

$$-12x-12y-39=y^{2}+4y-95-y^{2}$$

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

$$-12x-12y-39=4y-95$$

Add 12y to both sides.

$$-12x-39=4y-95+12y$$

Combine 4y and 12y to get 16y.

$$-12x-39=16y-95$$

Add 39 to both sides.

$$-12x=16y-95+39$$

Add -95 and 39 to get -56.

$$-12x=16y-56$$

Divide both sides by -12.

$$\frac{-12x}{-12}=\frac{16y-56}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x=\frac{16y-56}{-12}$$

Divide 16y-56 by -12.

$$x=\frac{14-4y}{3}$$

Solve for y

$$y=-\frac{3x}{4}+\frac{7}{2}$$

Steps for Solving Linear Equation

Subtract y^{2} from both sides.

$$x^{2}-10x+y^{2}-12y-39-y^{2}=x^{2}+2x+4y-95$$

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

$$x^{2}-10x-12y-39=x^{2}+2x+4y-95$$

Subtract 4y from both sides.

$$x^{2}-10x-12y-39-4y=x^{2}+2x-95$$

Combine -12y and -4y to get -16y.

$$x^{2}-10x-16y-39=x^{2}+2x-95$$

Subtract x^{2} from both sides.

$$-10x-16y-39=x^{2}+2x-95-x^{2}$$

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

$$-10x-16y-39=2x-95$$

Add 10x to both sides.

$$-16y-39=2x-95+10x$$

Combine 2x and 10x to get 12x.

$$-16y-39=12x-95$$

Add 39 to both sides.

$$-16y=12x-95+39$$

Add -95 and 39 to get -56.

$$-16y=12x-56$$

Divide both sides by -16.

$$\frac{-16y}{-16}=\frac{12x-56}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$y=\frac{12x-56}{-16}$$

Divide 12x-56 by -16.

$$y=-\frac{3x}{4}+\frac{7}{2}$$