Solve {l}{x^2+y^2=9}{x+y+1=0}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 9 } \\ { x + y + 1 = 0 } \end{array} \right.$$

Solve for x, y

$$x=\frac{\sqrt{17}-1}{2}\approx 1.561552813\text{, }y=\frac{-\sqrt{17}-1}{2}\approx -2.561552813$$

$$x=\frac{-\sqrt{17}-1}{2}\approx -2.561552813\text{, }y=\frac{\sqrt{17}-1}{2}\approx 1.561552813$$

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.

$$x+y+1=0,y^{2}+x^{2}=9$$

Solve x+y+1=0 for x by isolating x on the left hand side of the equal sign.

$$x+y+1=0$$

Subtract 1 from both sides of the equation.

$$x+y=-1$$

Subtract y from both sides of the equation.

$$x=-y-1$$

Substitute -y-1 for x in the other equation, y^{2}+x^{2}=9.

$$y^{2}+\left(-y-1\right)^{2}=9$$

Square -y-1.

$$y^{2}+y^{2}+2y+1=9$$

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

$$2y^{2}+2y+1=9$$

Subtract 9 from both sides of the equation.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\left(-1\right)\left(-1\right)\times 2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

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

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

Multiply -8 times -8.

$$y=\frac{-2±\sqrt{4+64}}{2\times 2}$$

Add 4 to 64.

$$y=\frac{-2±\sqrt{68}}{2\times 2}$$

Take the square root of 68.

$$y=\frac{-2±2\sqrt{17}}{2\times 2}$$

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

$$y=\frac{-2±2\sqrt{17}}{4}$$

Now solve the equation y=\frac{-2±2\sqrt{17}}{4} when ± is plus. Add -2 to 2\sqrt{17}.

$$y=\frac{2\sqrt{17}-2}{4}$$

Divide -2+2\sqrt{17} by 4.

$$y=\frac{\sqrt{17}-1}{2}$$

Now solve the equation y=\frac{-2±2\sqrt{17}}{4} when ± is minus. Subtract 2\sqrt{17} from -2.

$$y=\frac{-2\sqrt{17}-2}{4}$$

Divide -2-2\sqrt{17} by 4.

$$y=\frac{-\sqrt{17}-1}{2}$$

There are two solutions for y: \frac{-1+\sqrt{17}}{2} and \frac{-1-\sqrt{17}}{2}. Substitute \frac{-1+\sqrt{17}}{2} for y in the equation x=-y-1 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{\sqrt{17}-1}{2}-1$$

Now substitute \frac{-1-\sqrt{17}}{2} for y in the equation x=-y-1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{-\sqrt{17}-1}{2}-1$$

The system is now solved.

$$x=-\frac{\sqrt{17}-1}{2}-1,y=\frac{\sqrt{17}-1}{2}\text{ or }x=-\frac{-\sqrt{17}-1}{2}-1,y=\frac{-\sqrt{17}-1}{2}$$