Solve {l}{x^2+y^2=9}{x+y=2}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 9 } \\ { x + y = 2 } \end{array} \right.$$

Solve for x, y

$$x=\frac{\sqrt{14}}{2}+1\approx 2.870828693\text{, }y=-\frac{\sqrt{14}}{2}+1\approx -0.870828693$$

$$x=-\frac{\sqrt{14}}{2}+1\approx -0.870828693\text{, }y=\frac{\sqrt{14}}{2}+1\approx 2.870828693$$

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.

$$x+y=2,y^{2}+x^{2}=9$$

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}=9.

$$y^{2}+\left(-y+2\right)^{2}=9$$

Square -y+2.

$$y^{2}+y^{2}-4y+4=9$$

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

$$2y^{2}-4y+4=9$$

Subtract 9 from both sides of the equation.

$$2y^{2}-4y-5=0$$

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

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

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

$$y=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-5\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-4\right)±\sqrt{16-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$y=\frac{-\left(-4\right)±\sqrt{16+40}}{2\times 2}$$

Add 16 to 40.

$$y=\frac{-\left(-4\right)±\sqrt{56}}{2\times 2}$$

Take the square root of 56.

$$y=\frac{-\left(-4\right)±2\sqrt{14}}{2\times 2}$$

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

$$y=\frac{4±2\sqrt{14}}{2\times 2}$$

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

$$y=\frac{4±2\sqrt{14}}{4}$$

Now solve the equation y=\frac{4±2\sqrt{14}}{4} when ± is plus. Add 4 to 2\sqrt{14}.

$$y=\frac{2\sqrt{14}+4}{4}$$

Divide 4+2\sqrt{14} by 4.

$$y=\frac{\sqrt{14}}{2}+1$$

Now solve the equation y=\frac{4±2\sqrt{14}}{4} when ± is minus. Subtract 2\sqrt{14} from 4.

$$y=\frac{4-2\sqrt{14}}{4}$$

Divide 4-2\sqrt{14} by 4.

$$y=-\frac{\sqrt{14}}{2}+1$$

There are two solutions for y: 1+\frac{\sqrt{14}}{2} and 1-\frac{\sqrt{14}}{2}. Substitute 1+\frac{\sqrt{14}}{2} for y in the equation x=-y+2 to find the corresponding solution for x that satisfies both equations.

$$x=-\left(\frac{\sqrt{14}}{2}+1\right)+2$$

Now substitute 1-\frac{\sqrt{14}}{2} for y in the equation x=-y+2 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\left(-\frac{\sqrt{14}}{2}+1\right)+2$$

The system is now solved.

$$x=-\left(\frac{\sqrt{14}}{2}+1\right)+2,y=\frac{\sqrt{14}}{2}+1\text{ or }x=-\left(-\frac{\sqrt{14}}{2}+1\right)+2,y=-\frac{\sqrt{14}}{2}+1$$