Solve {l}{4x^2-2y^2=-2}{2x=1/3y+3}

$$\left\{ \begin{array} { l } { 4 x ^ { 2 } - 2 y ^ { 2 } = - 2 } \\ { 2 x = \frac { 1 } { 3 } y + 3 } \end{array} \right.$$

Solve for x, y

$$x=\frac{20}{17}\approx 1.176470588\text{, }y=-\frac{33}{17}\approx -1.941176471$$

$$x=2\text{, }y=3$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract \frac{1}{3}y from both sides.

$$2x-\frac{1}{3}y=3$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

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

Solve 2x-\frac{1}{3}y=3 for x by isolating x on the left hand side of the equal sign.

$$2x-\frac{1}{3}y=3$$

Subtract -\frac{1}{3}y from both sides of the equation.

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

Divide both sides by 2.

$$x=\frac{1}{6}y+\frac{3}{2}$$

Substitute \frac{1}{6}y+\frac{3}{2} for x in the other equation, -2y^{2}+4x^{2}=-2.

$$-2y^{2}+4\left(\frac{1}{6}y+\frac{3}{2}\right)^{2}=-2$$

Square \frac{1}{6}y+\frac{3}{2}.

$$-2y^{2}+4\left(\frac{1}{36}y^{2}+\frac{1}{2}y+\frac{9}{4}\right)=-2$$

Multiply 4 times \frac{1}{36}y^{2}+\frac{1}{2}y+\frac{9}{4}.

$$-2y^{2}+\frac{1}{9}y^{2}+2y+9=-2$$

Add -2y^{2} to \frac{1}{9}y^{2}.

$$-\frac{17}{9}y^{2}+2y+9=-2$$

Add 2 to both sides of the equation.

$$-\frac{17}{9}y^{2}+2y+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2+4\times \left(\frac{1}{6}\right)^{2} for a, 4\times \frac{3}{2}\times \frac{1}{6}\times 2 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-2±\sqrt{2^{2}-4\left(-\frac{17}{9}\right)\times 11}}{2\left(-\frac{17}{9}\right)}$$

Square 4\times \frac{3}{2}\times \frac{1}{6}\times 2.

$$y=\frac{-2±\sqrt{4-4\left(-\frac{17}{9}\right)\times 11}}{2\left(-\frac{17}{9}\right)}$$

Multiply -4 times -2+4\times \left(\frac{1}{6}\right)^{2}.

$$y=\frac{-2±\sqrt{4+\frac{68}{9}\times 11}}{2\left(-\frac{17}{9}\right)}$$

Multiply \frac{68}{9} times 11.

$$y=\frac{-2±\sqrt{4+\frac{748}{9}}}{2\left(-\frac{17}{9}\right)}$$

Add 4 to \frac{748}{9}.

$$y=\frac{-2±\sqrt{\frac{784}{9}}}{2\left(-\frac{17}{9}\right)}$$

Take the square root of \frac{784}{9}.

$$y=\frac{-2±\frac{28}{3}}{2\left(-\frac{17}{9}\right)}$$

Multiply 2 times -2+4\times \left(\frac{1}{6}\right)^{2}.

$$y=\frac{-2±\frac{28}{3}}{-\frac{34}{9}}$$

Now solve the equation y=\frac{-2±\frac{28}{3}}{-\frac{34}{9}} when ± is plus. Add -2 to \frac{28}{3}.

$$y=\frac{\frac{22}{3}}{-\frac{34}{9}}$$

Divide \frac{22}{3} by -\frac{34}{9} by multiplying \frac{22}{3} by the reciprocal of -\frac{34}{9}.

$$y=-\frac{33}{17}$$

Now solve the equation y=\frac{-2±\frac{28}{3}}{-\frac{34}{9}} when ± is minus. Subtract \frac{28}{3} from -2.

$$y=-\frac{\frac{34}{3}}{-\frac{34}{9}}$$

Divide -\frac{34}{3} by -\frac{34}{9} by multiplying -\frac{34}{3} by the reciprocal of -\frac{34}{9}.

$$y=3$$

There are two solutions for y: -\frac{33}{17} and 3. Substitute -\frac{33}{17} for y in the equation x=\frac{1}{6}y+\frac{3}{2} to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{6}\left(-\frac{33}{17}\right)+\frac{3}{2}$$

Multiply \frac{1}{6} times -\frac{33}{17} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{11}{34}+\frac{3}{2}$$

Add -\frac{33}{17}\times \frac{1}{6} to \frac{3}{2}.

$$x=\frac{20}{17}$$

Now substitute 3 for y in the equation x=\frac{1}{6}y+\frac{3}{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{6}\times 3+\frac{3}{2}$$

Multiply \frac{1}{6} times 3.

$$x=\frac{1+3}{2}$$

Add \frac{1}{6}\times 3 to \frac{3}{2}.

$$x=2$$

The system is now solved.

$$x=\frac{20}{17},y=-\frac{33}{17}\text{ or }x=2,y=3$$