Solve y^2+6y+3y+18=y^2+2y+8y+10

$${ y }^{ 2 } +6y+3y+18 = { y }^{ 2 } +2y+8y+10$$

Solve for y

$$y=8$$

Steps for Solving Linear Equation

Combine 6y and 3y to get 9y.

$$y^{2}+9y+18=y^{2}+2y+8y+10$$

Combine 2y and 8y to get 10y.

$$y^{2}+9y+18=y^{2}+10y+10$$

Subtract y^{2} from both sides.

$$y^{2}+9y+18-y^{2}=10y+10$$

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

$$9y+18=10y+10$$

Subtract 10y from both sides.

$$9y+18-10y=10$$

Combine 9y and -10y to get -y.

$$-y+18=10$$

Subtract 18 from both sides.

$$-y=10-18$$

Subtract 18 from 10 to get -8.

$$-y=-8$$

Multiply both sides by -1.

$$y=8$$