Solve {l}{x+y=1}{x^2+y^2=3}

$$\left\{ \begin{array} { l } { x + y = 1 } \\ { x ^ { 2 } + y ^ { 2 } = 3 } \end{array} \right.$$

Solve for x, y

$$x=\frac{\sqrt{5}+1}{2}\approx 1.618033989\text{, }y=\frac{1-\sqrt{5}}{2}\approx -0.618033989$$

$$x=\frac{1-\sqrt{5}}{2}\approx -0.618033989\text{, }y=\frac{\sqrt{5}+1}{2}\approx 1.618033989$$

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.

$$x+y=1,y^{2}+x^{2}=3$$

Solve x+y=1 for x by isolating x on the left hand side of the equal sign.

$$x+y=1$$

Subtract y from both sides of the equation.

$$x=-y+1$$

Substitute -y+1 for x in the other equation, y^{2}+x^{2}=3.

$$y^{2}+\left(-y+1\right)^{2}=3$$

Square -y+1.

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

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

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

Subtract 3 from both sides of the equation.

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

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

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

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

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

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

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

Multiply -8 times -2.

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

Add 4 to 16.

$$y=\frac{-\left(-2\right)±\sqrt{20}}{2\times 2}$$

Take the square root of 20.

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

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

$$y=\frac{2±2\sqrt{5}}{2\times 2}$$

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

$$y=\frac{2±2\sqrt{5}}{4}$$

Now solve the equation y=\frac{2±2\sqrt{5}}{4} when ± is plus. Add 2 to 2\sqrt{5}.

$$y=\frac{2\sqrt{5}+2}{4}$$

Divide 2+2\sqrt{5} by 4.

$$y=\frac{\sqrt{5}+1}{2}$$

Now solve the equation y=\frac{2±2\sqrt{5}}{4} when ± is minus. Subtract 2\sqrt{5} from 2.

$$y=\frac{2-2\sqrt{5}}{4}$$

Divide 2-2\sqrt{5} by 4.

$$y=\frac{1-\sqrt{5}}{2}$$

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

$$x=-\frac{\sqrt{5}+1}{2}+1$$

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

$$x=-\frac{1-\sqrt{5}}{2}+1$$

The system is now solved.

$$x=-\frac{\sqrt{5}+1}{2}+1,y=\frac{\sqrt{5}+1}{2}\text{ or }x=-\frac{1-\sqrt{5}}{2}+1,y=\frac{1-\sqrt{5}}{2}$$