Solve {l}{x^2+y^2=18}{x-y=6}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 18 } \\ { x - y = 6 } \end{array} \right.$$

Solve for x, y

$$x=3$$

$$y=-3$$

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.

$$x-y=6,y^{2}+x^{2}=18$$

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}=18.

$$y^{2}+\left(y+6\right)^{2}=18$$

Square y+6.

$$y^{2}+y^{2}+12y+36=18$$

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

$$2y^{2}+12y+36=18$$

Subtract 18 from both sides of the equation.

$$2y^{2}+12y+18=0$$

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

$$y=\frac{-12±\sqrt{12^{2}-4\times 2\times 18}}{2\times 2}$$

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

$$y=\frac{-12±\sqrt{144-4\times 2\times 18}}{2\times 2}$$

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

$$y=\frac{-12±\sqrt{144-8\times 18}}{2\times 2}$$

Multiply -8 times 18.

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

Add 144 to -144.

$$y=\frac{-12±\sqrt{0}}{2\times 2}$$

Take the square root of 0.

$$y=-\frac{12}{2\times 2}$$

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

$$y=-\frac{12}{4}$$

Divide -12 by 4.

$$y=-3$$

There are two solutions for y: -3 and -3. Substitute -3 for y in the equation x=y+6 to find the corresponding solution for x that satisfies both equations.

$$x=-3+6$$

Add -3 to 6.

$$x=3$$

The system is now solved.

$$x=3,y=-3\text{ or }x=3,y=-3$$