Solve {l}{x^2+y^2=10}{x-y=-2=10}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 10 } \\ { x - y = - 2 = 10 } \end{array} \right.$$

Solve for x, y

$$x=1\text{, }y=3$$

$$x=-3\text{, }y=-1$$

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.

$$x-y=-2,y^{2}+x^{2}=10$$

Solve x-y=-2 for x by isolating x on the left hand side of the equal sign.

$$x-y=-2$$

Subtract -y from both sides of the equation.

$$x=y-2$$

Substitute y-2 for x in the other equation, y^{2}+x^{2}=10.

$$y^{2}+\left(y-2\right)^{2}=10$$

Square y-2.

$$y^{2}+y^{2}-4y+4=10$$

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

$$2y^{2}-4y+4=10$$

Subtract 10 from both sides of the equation.

$$2y^{2}-4y-6=0$$

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

$$y=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-6\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-4\right)±\sqrt{16-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$y=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 2}$$

Add 16 to 48.

$$y=\frac{-\left(-4\right)±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$y=\frac{-\left(-4\right)±8}{2\times 2}$$

The opposite of 1\left(-2\right)\times 1\times 2 is 4.

$$y=\frac{4±8}{2\times 2}$$

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

$$y=\frac{4±8}{4}$$

Now solve the equation y=\frac{4±8}{4} when ± is plus. Add 4 to 8.

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

Divide 12 by 4.

$$y=3$$

Now solve the equation y=\frac{4±8}{4} when ± is minus. Subtract 8 from 4.

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

Divide -4 by 4.

$$y=-1$$

There are two solutions for y: 3 and -1. Substitute 3 for y in the equation x=y-2 to find the corresponding solution for x that satisfies both equations.

$$x=3-2$$

Add 1\times 3 to -2.

$$x=1$$

Now substitute -1 for y in the equation x=y-2 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-1-2$$

Add -1 to -2.

$$x=-3$$

The system is now solved.

$$x=1,y=3\text{ or }x=-3,y=-1$$