Solve {l}{x^2+y^2=116}{x-y=6}

$$\left. \begin{array}{l}{ x ^ { 2 } + y ^ { 2 } = 116 }\\{ x - y = 6 }\end{array} \right.$$

Solve for x, y

$$x=10\text{, }y=4$$

$$x=-4\text{, }y=-10$$

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.

$$x-y=6,y^{2}+x^{2}=116$$

Solve x-y=6 for x by isolating x on the left hand side of the equal sign.

$$x-y=6$$

Subtract -y from both sides of the equation.

$$x=y+6$$

Substitute y+6 for x in the other equation, y^{2}+x^{2}=116.

$$y^{2}+\left(y+6\right)^{2}=116$$

Square y+6.

$$y^{2}+y^{2}+12y+36=116$$

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

$$2y^{2}+12y+36=116$$

Subtract 116 from both sides of the equation.

$$2y^{2}+12y-80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 6\times 1\times 2 for b, and -80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-12±\sqrt{12^{2}-4\times 2\left(-80\right)}}{2\times 2}$$

Square 1\times 6\times 1\times 2.

$$y=\frac{-12±\sqrt{144-4\times 2\left(-80\right)}}{2\times 2}$$

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

$$y=\frac{-12±\sqrt{144-8\left(-80\right)}}{2\times 2}$$

Multiply -8 times -80.

$$y=\frac{-12±\sqrt{144+640}}{2\times 2}$$

Add 144 to 640.

$$y=\frac{-12±\sqrt{784}}{2\times 2}$$

Take the square root of 784.

$$y=\frac{-12±28}{2\times 2}$$

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

$$y=\frac{-12±28}{4}$$

Now solve the equation y=\frac{-12±28}{4} when ± is plus. Add -12 to 28.

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

Divide 16 by 4.

$$y=4$$

Now solve the equation y=\frac{-12±28}{4} when ± is minus. Subtract 28 from -12.

$$y=-\frac{40}{4}$$

Divide -40 by 4.

$$y=-10$$

There are two solutions for y: 4 and -10. Substitute 4 for y in the equation x=y+6 to find the corresponding solution for x that satisfies both equations.

$$x=4+6$$

Add 1\times 4 to 6.

$$x=10$$

Now substitute -10 for y in the equation x=y+6 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-10+6$$

Add -10 to 6.

$$x=-4$$

The system is now solved.

$$x=10,y=4\text{ or }x=-4,y=-10$$