Solve {l}{x^2+y^2=6}{y+x=1}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 6 } \\ { y + x = 1 } \end{array} \right.$$

Solve for x, y

$$x=\frac{\sqrt{11}+1}{2}\approx 2.158312395\text{, }y=\frac{1-\sqrt{11}}{2}\approx -1.158312395$$

$$x=\frac{1-\sqrt{11}}{2}\approx -1.158312395\text{, }y=\frac{\sqrt{11}+1}{2}\approx 2.158312395$$

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.

$$y+x=1,x^{2}+y^{2}=6$$

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

$$y+x=1$$

Subtract x from both sides of the equation.

$$y=-x+1$$

Substitute -x+1 for y in the other equation, x^{2}+y^{2}=6.

$$x^{2}+\left(-x+1\right)^{2}=6$$

Square -x+1.

$$x^{2}+x^{2}-2x+1=6$$

Add x^{2} to x^{2}.

$$2x^{2}-2x+1=6$$

Subtract 6 from both sides of the equation.

$$2x^{2}-2x-5=0$$

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 -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

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

$$x=\frac{-\left(-2\right)±\sqrt{4-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$x=\frac{-\left(-2\right)±\sqrt{4+40}}{2\times 2}$$

Add 4 to 40.

$$x=\frac{-\left(-2\right)±\sqrt{44}}{2\times 2}$$

Take the square root of 44.

$$x=\frac{-\left(-2\right)±2\sqrt{11}}{2\times 2}$$

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

$$x=\frac{2±2\sqrt{11}}{2\times 2}$$

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

$$x=\frac{2±2\sqrt{11}}{4}$$

Now solve the equation x=\frac{2±2\sqrt{11}}{4} when ± is plus. Add 2 to 2\sqrt{11}.

$$x=\frac{2\sqrt{11}+2}{4}$$

Divide 2+2\sqrt{11} by 4.

$$x=\frac{\sqrt{11}+1}{2}$$

Now solve the equation x=\frac{2±2\sqrt{11}}{4} when ± is minus. Subtract 2\sqrt{11} from 2.

$$x=\frac{2-2\sqrt{11}}{4}$$

Divide 2-2\sqrt{11} by 4.

$$x=\frac{1-\sqrt{11}}{2}$$

There are two solutions for x: \frac{1+\sqrt{11}}{2} and \frac{1-\sqrt{11}}{2}. Substitute \frac{1+\sqrt{11}}{2} for x in the equation y=-x+1 to find the corresponding solution for y that satisfies both equations.

$$y=-\frac{\sqrt{11}+1}{2}+1$$

Now substitute \frac{1-\sqrt{11}}{2} for x in the equation y=-x+1 and solve to find the corresponding solution for y that satisfies both equations.

$$y=-\frac{1-\sqrt{11}}{2}+1$$

The system is now solved.

$$y=-\frac{\sqrt{11}+1}{2}+1,x=\frac{\sqrt{11}+1}{2}\text{ or }y=-\frac{1-\sqrt{11}}{2}+1,x=\frac{1-\sqrt{11}}{2}$$