Solve {l}{x^2+2y^2=3}{x+y=2}

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

Solve for x, y

$$x=\frac{5}{3}\approx 1.666666667\text{, }y=\frac{1}{3}\approx 0.333333333$$

$$x=1\text{, }y=1$$

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.

$$x+y=2,2y^{2}+x^{2}=3$$

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

$$x+y=2$$

Subtract y from both sides of the equation.

$$x=-y+2$$

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

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

Square -y+2.

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

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

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

Subtract 3 from both sides of the equation.

$$3y^{2}-4y+1=0$$

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

$$y=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 3}}{2\times 3}$$

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

$$y=\frac{-\left(-4\right)±\sqrt{16-4\times 3}}{2\times 3}$$

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

$$y=\frac{-\left(-4\right)±\sqrt{16-12}}{2\times 3}$$

Add 16 to -12.

$$y=\frac{-\left(-4\right)±\sqrt{4}}{2\times 3}$$

Take the square root of 4.

$$y=\frac{-\left(-4\right)±2}{2\times 3}$$

The opposite of 1\times 2\left(-1\right)\times 2 is 4.

$$y=\frac{4±2}{2\times 3}$$

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

$$y=\frac{4±2}{6}$$

Now solve the equation y=\frac{4±2}{6} when ± is plus. Add 4 to 2.

$$y=\frac{6}{6}$$

Divide 6 by 6.

$$y=1$$

Now solve the equation y=\frac{4±2}{6} when ± is minus. Subtract 2 from 4.

$$y=\frac{2}{6}$$

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

$$y=\frac{1}{3}$$

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

$$x=-1+2$$

Add -1 to 2.

$$x=1$$

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

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

Add -\frac{1}{3} to 2.

$$x=\frac{5}{3}$$

The system is now solved.

$$x=1,y=1\text{ or }x=\frac{5}{3},y=\frac{1}{3}$$