Solve {l}{x+y=2}{x^2+y^2=12}

$$\left. \begin{array} { l } { x + y = 2 } \\ { x ^ { 2 } + y ^ { 2 } = 12 } \end{array} \right.$$

Solve for x, y

$$x=\sqrt{5}+1\approx 3.236067977\text{, }y=1-\sqrt{5}\approx -1.236067977$$

$$x=1-\sqrt{5}\approx -1.236067977\text{, }y=\sqrt{5}+1\approx 3.236067977$$

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.

$$x+y=2,y^{2}+x^{2}=12$$

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, y^{2}+x^{2}=12.

$$y^{2}+\left(-y+2\right)^{2}=12$$

Square -y+2.

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

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

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

Subtract 12 from both sides of the equation.

$$2y^{2}-4y-8=0$$

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 -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

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

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

Multiply -8 times -8.

$$y=\frac{-\left(-4\right)±\sqrt{16+64}}{2\times 2}$$

Add 16 to 64.

$$y=\frac{-\left(-4\right)±\sqrt{80}}{2\times 2}$$

Take the square root of 80.

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

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

$$y=\frac{4±4\sqrt{5}}{2\times 2}$$

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

$$y=\frac{4±4\sqrt{5}}{4}$$

Now solve the equation y=\frac{4±4\sqrt{5}}{4} when ± is plus. Add 4 to 4\sqrt{5}.

$$y=\frac{4\sqrt{5}+4}{4}$$

Divide 4+4\sqrt{5} by 4.

$$y=\sqrt{5}+1$$

Now solve the equation y=\frac{4±4\sqrt{5}}{4} when ± is minus. Subtract 4\sqrt{5} from 4.

$$y=\frac{4-4\sqrt{5}}{4}$$

Divide 4-4\sqrt{5} by 4.

$$y=1-\sqrt{5}$$

There are two solutions for y: 1+\sqrt{5} and 1-\sqrt{5}. Substitute 1+\sqrt{5} for y in the equation x=-y+2 to find the corresponding solution for x that satisfies both equations.

$$x=-\left(\sqrt{5}+1\right)+2$$

Now substitute 1-\sqrt{5} for y in the equation x=-y+2 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\left(1-\sqrt{5}\right)+2$$

The system is now solved.

$$x=-\left(\sqrt{5}+1\right)+2,y=\sqrt{5}+1\text{ or }x=-\left(1-\sqrt{5}\right)+2,y=1-\sqrt{5}$$