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

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

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.

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

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

$$x+y=1$$

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\times 1\left(-1\right)\times 2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-2\right)^{2}.

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

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

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

Divide 4 by 4.

$$y=1$$

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

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

Divide 0 by 4.

$$y=0$$

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

$$x=-1+1$$

Add -1 to 1.

$$x=0$$

Now substitute 0 for y in the equation x=-y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=1$$

The system is now solved.

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