Solve {l}{text{A.}}{x^2+2y^2+3z^2-2x+4y=0}

$$\left. \begin{array} { l } { \text { A. } } \\ { x ^ { 2 } + 2 y ^ { 2 } + 3 z ^ { 2 } - 2 x + 4 y = 0 } \end{array} \right.$$

Solve for x

$$x=\frac{3z^{2}}{2}+y^{2}+2y$$

Steps for Solving Linear Equation

Subtract 2y^{2} from both sides. Anything subtracted from zero gives its negation.

$$3z^{2}-2x+4y=-2y^{2}$$

Subtract 3z^{2} from both sides.

$$-2x+4y=-2y^{2}-3z^{2}$$

Subtract 4y from both sides.

$$-2x=-2y^{2}-3z^{2}-4y$$

The equation is in standard form.

$$-2x=-2y^{2}-4y-3z^{2}$$

Divide both sides by -2.

$$\frac{-2x}{-2}=\frac{-2y^{2}-4y-3z^{2}}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x=\frac{-2y^{2}-4y-3z^{2}}{-2}$$

Divide -2y^{2}-3z^{2}-4y by -2.

$$x=\frac{3z^{2}}{2}+y^{2}+2y$$