Solve {l}{8=11(x-y)}{32=11(x^2+y^2)}

$$\left\{ \begin{array} { l } { 8 = 11 ( x - y ) } \\ { 32 = 11 ( x ^ { 2 } + y ^ { 2 } ) } \end{array} \right.$$

Solve for x, y

$$x=\frac{4-4\sqrt{10}}{11}\approx -0.786282786\text{, }y=\frac{-4\sqrt{10}-4}{11}\approx -1.513555513$$

$$x=\frac{4\sqrt{10}+4}{11}\approx 1.513555513\text{, }y=\frac{4\sqrt{10}-4}{11}\approx 0.786282786$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Divide both sides by 11.

$$\frac{8}{11}=x-y$$

Swap sides so that all variable terms are on the left hand side.

$$x-y=\frac{8}{11}$$

Consider the second equation. Divide both sides by 11.

$$\frac{32}{11}=x^{2}+y^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+y^{2}=\frac{32}{11}$$

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.

$$x-y=\frac{8}{11},y^{2}+x^{2}=\frac{32}{11}$$

Solve x-y=\frac{8}{11} for x by isolating x on the left hand side of the equal sign.

$$x-y=\frac{8}{11}$$

Subtract -y from both sides of the equation.

$$x=y+\frac{8}{11}$$

Substitute y+\frac{8}{11} for x in the other equation, y^{2}+x^{2}=\frac{32}{11}.

$$y^{2}+\left(y+\frac{8}{11}\right)^{2}=\frac{32}{11}$$

Square y+\frac{8}{11}.

$$y^{2}+y^{2}+\frac{16}{11}y+\frac{64}{121}=\frac{32}{11}$$

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

$$2y^{2}+\frac{16}{11}y+\frac{64}{121}=\frac{32}{11}$$

Subtract \frac{32}{11} from both sides of the equation.

$$2y^{2}+\frac{16}{11}y-\frac{288}{121}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times \frac{8}{11}\times 1\times 2 for b, and -\frac{288}{121} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\frac{16}{11}±\sqrt{\left(\frac{16}{11}\right)^{2}-4\times 2\left(-\frac{288}{121}\right)}}{2\times 2}$$

Square 1\times \frac{8}{11}\times 1\times 2.

$$y=\frac{-\frac{16}{11}±\sqrt{\frac{256}{121}-4\times 2\left(-\frac{288}{121}\right)}}{2\times 2}$$

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

$$y=\frac{-\frac{16}{11}±\sqrt{\frac{256}{121}-8\left(-\frac{288}{121}\right)}}{2\times 2}$$

Multiply -8 times -\frac{288}{121}.

$$y=\frac{-\frac{16}{11}±\sqrt{\frac{256+2304}{121}}}{2\times 2}$$

Add \frac{256}{121} to \frac{2304}{121} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{16}{11}±\sqrt{\frac{2560}{121}}}{2\times 2}$$

Take the square root of \frac{2560}{121}.

$$y=\frac{-\frac{16}{11}±\frac{16\sqrt{10}}{11}}{2\times 2}$$

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

$$y=\frac{-\frac{16}{11}±\frac{16\sqrt{10}}{11}}{4}$$

Now solve the equation y=\frac{-\frac{16}{11}±\frac{16\sqrt{10}}{11}}{4} when ± is plus. Add -\frac{16}{11} to \frac{16\sqrt{10}}{11}.

$$y=\frac{16\sqrt{10}-16}{4\times 11}$$

Divide \frac{-16+16\sqrt{10}}{11} by 4.

$$y=\frac{4\sqrt{10}-4}{11}$$

Now solve the equation y=\frac{-\frac{16}{11}±\frac{16\sqrt{10}}{11}}{4} when ± is minus. Subtract \frac{16\sqrt{10}}{11} from -\frac{16}{11}.

$$y=\frac{-16\sqrt{10}-16}{4\times 11}$$

Divide \frac{-16-16\sqrt{10}}{11} by 4.

$$y=\frac{-4\sqrt{10}-4}{11}$$

There are two solutions for y: \frac{-4+4\sqrt{10}}{11} and \frac{-4-4\sqrt{10}}{11}. Substitute \frac{-4+4\sqrt{10}}{11} for y in the equation x=y+\frac{8}{11} to find the corresponding solution for x that satisfies both equations.

$$x=\frac{4\sqrt{10}-4+8}{11}$$

Now substitute \frac{-4-4\sqrt{10}}{11} for y in the equation x=y+\frac{8}{11} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{-4\sqrt{10}-4+8}{11}$$

The system is now solved.

$$x=\frac{4\sqrt{10}-4+8}{11},y=\frac{4\sqrt{10}-4}{11}\text{ or }x=\frac{-4\sqrt{10}-4+8}{11},y=\frac{-4\sqrt{10}-4}{11}$$