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

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

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

Consider the second equation. Subtract x from both sides.

$$y-x=2$$

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.

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

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

$$y-x=2$$

Subtract -x from both sides of the equation.

$$y=x+2$$

Substitute x+2 for y in the other equation, x^{2}+y^{2}=9.

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

Square x+2.

$$x^{2}+x^{2}+4x+4=9$$

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

$$2x^{2}+4x+4=9$$

Subtract 9 from both sides of the equation.

$$2x^{2}+4x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 2\times 1\times 2 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 1\times 2\times 1\times 2.

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

Multiply -4 times 1+1\times 1^{2}.

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

Multiply -8 times -5.

$$x=\frac{-4±\sqrt{16+40}}{2\times 2}$$

Add 16 to 40.

$$x=\frac{-4±\sqrt{56}}{2\times 2}$$

Take the square root of 56.

$$x=\frac{-4±2\sqrt{14}}{2\times 2}$$

Multiply 2 times 1+1\times 1^{2}.

$$x=\frac{-4±2\sqrt{14}}{4}$$

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

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

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

$$x=\frac{\sqrt{14}}{2}-1$$

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

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

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

$$x=-\frac{\sqrt{14}}{2}-1$$

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

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

Add 1\left(-1+\frac{\sqrt{14}}{2}\right) to 2.

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

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

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

Add 1\left(-1-\frac{\sqrt{14}}{2}\right) to 2.

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

The system is now solved.

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