Solve {l}{x+y=3}{x^2+2y^2=6}

$$\left\{ \begin{array} { l } { x + y = 3 } \\ { x ^ { 2 } + 2 y ^ { 2 } = 6 } \end{array} \right.$$

Solve for x, y

$$x=2$$

$$y=1$$

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.

$$x+y=3,2y^{2}+x^{2}=6$$

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

$$x+y=3$$

Subtract y from both sides of the equation.

$$x=-y+3$$

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

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

Square -y+3.

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

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

$$3y^{2}-6y+9=6$$

Subtract 6 from both sides of the equation.

$$3y^{2}-6y+3=0$$

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

$$y=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3\times 3}}{2\times 3}$$

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

$$y=\frac{-\left(-6\right)±\sqrt{36-4\times 3\times 3}}{2\times 3}$$

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

$$y=\frac{-\left(-6\right)±\sqrt{36-12\times 3}}{2\times 3}$$

Multiply -12 times 3.

$$y=\frac{-\left(-6\right)±\sqrt{36-36}}{2\times 3}$$

Add 36 to -36.

$$y=\frac{-\left(-6\right)±\sqrt{0}}{2\times 3}$$

Take the square root of 0.

$$y=-\frac{-6}{2\times 3}$$

The opposite of 1\times 3\left(-1\right)\times 2 is 6.

$$y=\frac{6}{2\times 3}$$

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

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

Divide 6 by 6.

$$y=1$$

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

$$x=-1+3$$

Add -1 to 3.

$$x=2$$

The system is now solved.

$$x=2,y=1\text{ or }x=2,y=1$$