Solve x^2/4+y/3=1

$$\frac { x ^ { 2 } } { 4 } + \frac { y } { 3 } = 1$$

Solve for y

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

Steps for Solving Linear Equation

Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$3x^{2}+4y=12$$

Subtract 3x^{2} from both sides.

$$4y=12-3x^{2}$$

Divide both sides by 4.

$$\frac{4y}{4}=\frac{12-3x^{2}}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y=\frac{12-3x^{2}}{4}$$

Divide -3x^{2}+12 by 4.

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

Solve for x (complex solution)

$$x=-\frac{2\sqrt{9-3y}}{3}$$

$$x=\frac{2\sqrt{9-3y}}{3}$$

Solve for x

$$x=\frac{2\sqrt{9-3y}}{3}$$

$$x=-\frac{2\sqrt{9-3y}}{3}\text{, }y\leq 3$$