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

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

Solve for x, y

$$x=-\frac{\sqrt{2}}{2}\approx -0.707106781\text{, }y=-\frac{\sqrt{2}}{2}\approx -0.707106781$$

$$x=\frac{\sqrt{2}}{2}\approx 0.707106781\text{, }y=\frac{\sqrt{2}}{2}\approx 0.707106781$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$x=y$$

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

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

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

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

Subtract 1 from both sides of the equation.

$$2y^{2}-1=0$$

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

$$y=\frac{0±\sqrt{0^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

Square 1\times 0\times 1\times 2.

$$y=\frac{0±\sqrt{-4\times 2\left(-1\right)}}{2\times 2}$$

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

$$y=\frac{0±\sqrt{-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$y=\frac{0±\sqrt{8}}{2\times 2}$$

Take the square root of 8.

$$y=\frac{0±2\sqrt{2}}{2\times 2}$$

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

$$y=\frac{0±2\sqrt{2}}{4}$$

Now solve the equation y=\frac{0±2\sqrt{2}}{4} when ± is plus.

$$y=\frac{\sqrt{2}}{2}$$

Now solve the equation y=\frac{0±2\sqrt{2}}{4} when ± is minus.

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

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

$$x=\frac{\sqrt{2}}{2}$$

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

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

The system is now solved.

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