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

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

Solve for x, y

$$x=0\text{, }y=-1$$

$$x=-1\text{, }y=0$$

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.

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

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}=1.

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

Square -y-1.

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

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

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

Subtract 1 from both sides of the equation.

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

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

$$y=\frac{-2±\sqrt{2^{2}}}{2\times 2}$$

Take the square root of 2^{2}.

$$y=\frac{-2±2}{2\times 2}$$

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

$$y=\frac{-2±2}{4}$$

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

$$y=\frac{0}{4}$$

Divide 0 by 4.

$$y=0$$

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

$$y=-\frac{4}{4}$$

Divide -4 by 4.

$$y=-1$$

There are two solutions for y: 0 and -1. Substitute 0 for y in the equation x=-y-1 to find the corresponding solution for x that satisfies both equations.

$$x=-1$$

Now substitute -1 for y in the equation x=-y-1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\left(-1\right)-1$$

Multiply -1 times -1.

$$x=1-1$$

Add -\left(-1\right) to -1.

$$x=0$$

The system is now solved.

$$x=-1,y=0\text{ or }x=0,y=-1$$