Solve {l}{x+2y-3=0}{x^2+2y^2=4}

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

Solve for x, y

$$x=\frac{\sqrt{6}}{3}+1\approx 1.816496581\text{, }y=-\frac{\sqrt{6}}{6}+1\approx 0.59175171$$

$$x=-\frac{\sqrt{6}}{3}+1\approx 0.183503419\text{, }y=\frac{\sqrt{6}}{6}+1\approx 1.40824829$$

Steps Using Substitution and Quadratic Formula

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.

$$x+2y-3=0,2y^{2}+x^{2}=4$$

Solve x+2y-3=0 for x by isolating x on the left hand side of the equal sign.

$$x+2y-3=0$$

Add 3 to both sides of the equation.

$$x+2y=3$$

Subtract 2y from both sides of the equation.

$$x=-2y+3$$

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

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

Square -2y+3.

$$2y^{2}+4y^{2}-12y+9=4$$

Add 2y^{2} to 4y^{2}.

$$6y^{2}-12y+9=4$$

Subtract 4 from both sides of the equation.

$$6y^{2}-12y+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2+1\left(-2\right)^{2} for a, 1\times 3\left(-2\right)\times 2 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 6\times 5}}{2\times 6}$$

Square 1\times 3\left(-2\right)\times 2.

$$y=\frac{-\left(-12\right)±\sqrt{144-4\times 6\times 5}}{2\times 6}$$

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

$$y=\frac{-\left(-12\right)±\sqrt{144-24\times 5}}{2\times 6}$$

Multiply -24 times 5.

$$y=\frac{-\left(-12\right)±\sqrt{144-120}}{2\times 6}$$

Add 144 to -120.

$$y=\frac{-\left(-12\right)±\sqrt{24}}{2\times 6}$$

Take the square root of 24.

$$y=\frac{-\left(-12\right)±2\sqrt{6}}{2\times 6}$$

The opposite of 1\times 3\left(-2\right)\times 2 is 12.

$$y=\frac{12±2\sqrt{6}}{2\times 6}$$

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

$$y=\frac{12±2\sqrt{6}}{12}$$

Now solve the equation y=\frac{12±2\sqrt{6}}{12} when ± is plus. Add 12 to 2\sqrt{6}.

$$y=\frac{2\sqrt{6}+12}{12}$$

Divide 12+2\sqrt{6} by 12.

$$y=\frac{\sqrt{6}}{6}+1$$

Now solve the equation y=\frac{12±2\sqrt{6}}{12} when ± is minus. Subtract 2\sqrt{6} from 12.

$$y=\frac{12-2\sqrt{6}}{12}$$

Divide 12-2\sqrt{6} by 12.

$$y=-\frac{\sqrt{6}}{6}+1$$

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

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

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

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

The system is now solved.

$$x=-2\left(\frac{\sqrt{6}}{6}+1\right)+3,y=\frac{\sqrt{6}}{6}+1\text{ or }x=-2\left(-\frac{\sqrt{6}}{6}+1\right)+3,y=-\frac{\sqrt{6}}{6}+1$$