Solve {l}{x-y=-11}{x^2+y^2=65}

$$\left. \begin{array}{l}{ x - y = - 11 }\\{ x ^ { 2 } + y ^ { 2 } = 65 }\end{array} \right.$$

Solve for x, y

$$x=-4\text{, }y=7$$

$$x=-7\text{, }y=4$$

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.

$$x-y=-11,y^{2}+x^{2}=65$$

Solve x-y=-11 for x by isolating x on the left hand side of the equal sign.

$$x-y=-11$$

Subtract -y from both sides of the equation.

$$x=y-11$$

Substitute y-11 for x in the other equation, y^{2}+x^{2}=65.

$$y^{2}+\left(y-11\right)^{2}=65$$

Square y-11.

$$y^{2}+y^{2}-22y+121=65$$

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

$$2y^{2}-22y+121=65$$

Subtract 65 from both sides of the equation.

$$2y^{2}-22y+56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\left(-11\right)\times 1\times 2 for b, and 56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 2\times 56}}{2\times 2}$$

Square 1\left(-11\right)\times 1\times 2.

$$y=\frac{-\left(-22\right)±\sqrt{484-4\times 2\times 56}}{2\times 2}$$

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

$$y=\frac{-\left(-22\right)±\sqrt{484-8\times 56}}{2\times 2}$$

Multiply -8 times 56.

$$y=\frac{-\left(-22\right)±\sqrt{484-448}}{2\times 2}$$

Add 484 to -448.

$$y=\frac{-\left(-22\right)±\sqrt{36}}{2\times 2}$$

Take the square root of 36.

$$y=\frac{-\left(-22\right)±6}{2\times 2}$$

The opposite of 1\left(-11\right)\times 1\times 2 is 22.

$$y=\frac{22±6}{2\times 2}$$

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

$$y=\frac{22±6}{4}$$

Now solve the equation y=\frac{22±6}{4} when ± is plus. Add 22 to 6.

$$y=\frac{28}{4}$$

Divide 28 by 4.

$$y=7$$

Now solve the equation y=\frac{22±6}{4} when ± is minus. Subtract 6 from 22.

$$y=\frac{16}{4}$$

Divide 16 by 4.

$$y=4$$

There are two solutions for y: 7 and 4. Substitute 7 for y in the equation x=y-11 to find the corresponding solution for x that satisfies both equations.

$$x=7-11$$

Add 1\times 7 to -11.

$$x=-4$$

Now substitute 4 for y in the equation x=y-11 and solve to find the corresponding solution for x that satisfies both equations.

$$x=4-11$$

Add 1\times 4 to -11.

$$x=-7$$

The system is now solved.

$$x=-4,y=7\text{ or }x=-7,y=4$$