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

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

Solve for x, y

$$x=1$$

$$y=1$$

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.

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

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

$$x+y-2=0$$

Add 2 to both sides of the equation.

$$x+y=2$$

Subtract y from both sides of the equation.

$$x=-y+2$$

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

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

Square -y+2.

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

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

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

Subtract 2 from both sides of the equation.

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

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

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

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

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

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

$$y=\frac{-\left(-4\right)±\sqrt{16-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

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

Add 16 to -16.

$$y=\frac{-\left(-4\right)±\sqrt{0}}{2\times 2}$$

Take the square root of 0.

$$y=-\frac{-4}{2\times 2}$$

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

$$y=\frac{4}{2\times 2}$$

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

$$y=\frac{4}{4}$$

Divide 4 by 4.

$$y=1$$

There are two solutions for y: 1 and 1. 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.

$$x=1$$

The system is now solved.

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