Solve {l}{x^2+y^2=6}{x-y=3}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 6 } \\ { x - y = 3 } \end{array} \right.$$

Solve for x, y

$$x=\frac{3-\sqrt{3}}{2}\approx 0.633974596\text{, }y=\frac{-\sqrt{3}-3}{2}\approx -2.366025404$$

$$x=\frac{\sqrt{3}+3}{2}\approx 2.366025404\text{, }y=\frac{\sqrt{3}-3}{2}\approx -0.633974596$$

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.

$$x-y=3,y^{2}+x^{2}=6$$

Solve x-y=3 for x by isolating x on the left hand side of the equal sign.

$$x-y=3$$

Subtract -y from both sides of the equation.

$$x=y+3$$

Substitute y+3 for x in the other equation, y^{2}+x^{2}=6.

$$y^{2}+\left(y+3\right)^{2}=6$$

Square y+3.

$$y^{2}+y^{2}+6y+9=6$$

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

$$2y^{2}+6y+9=6$$

Subtract 6 from both sides of the equation.

$$2y^{2}+6y+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 3\times 1\times 2 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-6±\sqrt{6^{2}-4\times 2\times 3}}{2\times 2}$$

Square 1\times 3\times 1\times 2.

$$y=\frac{-6±\sqrt{36-4\times 2\times 3}}{2\times 2}$$

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

$$y=\frac{-6±\sqrt{36-8\times 3}}{2\times 2}$$

Multiply -8 times 3.

$$y=\frac{-6±\sqrt{36-24}}{2\times 2}$$

Add 36 to -24.

$$y=\frac{-6±\sqrt{12}}{2\times 2}$$

Take the square root of 12.

$$y=\frac{-6±2\sqrt{3}}{2\times 2}$$

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

$$y=\frac{-6±2\sqrt{3}}{4}$$

Now solve the equation y=\frac{-6±2\sqrt{3}}{4} when ± is plus. Add -6 to 2\sqrt{3}.

$$y=\frac{2\sqrt{3}-6}{4}$$

Divide -6+2\sqrt{3} by 4.

$$y=\frac{\sqrt{3}-3}{2}$$

Now solve the equation y=\frac{-6±2\sqrt{3}}{4} when ± is minus. Subtract 2\sqrt{3} from -6.

$$y=\frac{-2\sqrt{3}-6}{4}$$

Divide -6-2\sqrt{3} by 4.

$$y=\frac{-\sqrt{3}-3}{2}$$

There are two solutions for y: \frac{-3+\sqrt{3}}{2} and \frac{-3-\sqrt{3}}{2}. Substitute \frac{-3+\sqrt{3}}{2} for y in the equation x=y+3 to find the corresponding solution for x that satisfies both equations.

$$x=\frac{\sqrt{3}-3}{2}+3$$

Now substitute \frac{-3-\sqrt{3}}{2} for y in the equation x=y+3 and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{-\sqrt{3}-3}{2}+3$$

The system is now solved.

$$x=\frac{\sqrt{3}-3}{2}+3,y=\frac{\sqrt{3}-3}{2}\text{ or }x=\frac{-\sqrt{3}-3}{2}+3,y=\frac{-\sqrt{3}-3}{2}$$