Solve {c}{x^2+y^2=16}{x=y}

$$\left. \begin{array} { c } { x ^ { 2 } + y ^ { 2 } = 16 } \\ { x = y } \end{array} \right.$$

Solve for x, y

$$x=-2\sqrt{2}\approx -2.828427125\text{, }y=-2\sqrt{2}\approx -2.828427125$$

$$x=2\sqrt{2}\approx 2.828427125\text{, }y=2\sqrt{2}\approx 2.828427125$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract y from both sides.

$$x-y=0$$

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.

$$x-y=0,y^{2}+x^{2}=16$$

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

$$x-y=0$$

Subtract -y from both sides of the equation.

$$x=y$$

Substitute y for x in the other equation, y^{2}+x^{2}=16.

$$y^{2}+y^{2}=16$$

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

$$2y^{2}=16$$

Subtract 16 from both sides of the equation.

$$2y^{2}-16=0$$

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 -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\times 2\left(-16\right)}}{2\times 2}$$

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

$$y=\frac{0±\sqrt{-4\times 2\left(-16\right)}}{2\times 2}$$

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

$$y=\frac{0±\sqrt{-8\left(-16\right)}}{2\times 2}$$

Multiply -8 times -16.

$$y=\frac{0±\sqrt{128}}{2\times 2}$$

Take the square root of 128.

$$y=\frac{0±8\sqrt{2}}{2\times 2}$$

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

$$y=\frac{0±8\sqrt{2}}{4}$$

Now solve the equation y=\frac{0±8\sqrt{2}}{4} when ± is plus.

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

Now solve the equation y=\frac{0±8\sqrt{2}}{4} when ± is minus.

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

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

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

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

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

The system is now solved.

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