Solve {l}{x^2+y^2=25}{y=x}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 25 } \\ { y = x } \end{array} \right.$$

Solve for x, y

$$x=-\frac{5\sqrt{2}}{2}\approx -3.535533906\text{, }y=-\frac{5\sqrt{2}}{2}\approx -3.535533906$$

$$x=\frac{5\sqrt{2}}{2}\approx 3.535533906\text{, }y=\frac{5\sqrt{2}}{2}\approx 3.535533906$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract x from both sides.

$$y-x=0$$

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.

$$y-x=0,x^{2}+y^{2}=25$$

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

$$y-x=0$$

Subtract -x from both sides of the equation.

$$y=x$$

Substitute x for y in the other equation, x^{2}+y^{2}=25.

$$x^{2}+x^{2}=25$$

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

$$2x^{2}=25$$

Subtract 25 from both sides of the equation.

$$2x^{2}-25=0$$

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 -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(-25\right)}}{2\times 2}$$

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

$$x=\frac{0±\sqrt{-4\times 2\left(-25\right)}}{2\times 2}$$

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

$$x=\frac{0±\sqrt{-8\left(-25\right)}}{2\times 2}$$

Multiply -8 times -25.

$$x=\frac{0±\sqrt{200}}{2\times 2}$$

Take the square root of 200.

$$x=\frac{0±10\sqrt{2}}{2\times 2}$$

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

$$x=\frac{0±10\sqrt{2}}{4}$$

Now solve the equation x=\frac{0±10\sqrt{2}}{4} when ± is plus.

$$x=\frac{5\sqrt{2}}{2}$$

Now solve the equation x=\frac{0±10\sqrt{2}}{4} when ± is minus.

$$x=-\frac{5\sqrt{2}}{2}$$

There are two solutions for x: \frac{5\sqrt{2}}{2} and -\frac{5\sqrt{2}}{2}. Substitute \frac{5\sqrt{2}}{2} for x in the equation y=x to find the corresponding solution for y that satisfies both equations.

$$y=\frac{5\sqrt{2}}{2}$$

Now substitute -\frac{5\sqrt{2}}{2} for x in the equation y=x and solve to find the corresponding solution for y that satisfies both equations.

$$y=-\frac{5\sqrt{2}}{2}$$

The system is now solved.

$$y=\frac{5\sqrt{2}}{2},x=\frac{5\sqrt{2}}{2}\text{ or }y=-\frac{5\sqrt{2}}{2},x=-\frac{5\sqrt{2}}{2}$$