Solve {l}{x-y=1}{x^2+y^2=13}

$$\left\{ \begin{array} { l } { x - y = 1 } \\ { x ^ { 2 } + y ^ { 2 } = 13 } \end{array} \right.$$

Solve for x, y

$$x=3\text{, }y=2$$

$$x=-2\text{, }y=-3$$

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.

$$x-y=1,y^{2}+x^{2}=13$$

Solve x-y=1 for x by isolating x on the left hand side of the equal sign.

$$x-y=1$$

Subtract -y from both sides of the equation.

$$x=y+1$$

Substitute y+1 for x in the other equation, y^{2}+x^{2}=13.

$$y^{2}+\left(y+1\right)^{2}=13$$

Square y+1.

$$y^{2}+y^{2}+2y+1=13$$

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

$$2y^{2}+2y+1=13$$

Subtract 13 from both sides of the equation.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 1\times 1\times 2 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

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

$$y=\frac{-2±\sqrt{4-8\left(-12\right)}}{2\times 2}$$

Multiply -8 times -12.

$$y=\frac{-2±\sqrt{4+96}}{2\times 2}$$

Add 4 to 96.

$$y=\frac{-2±\sqrt{100}}{2\times 2}$$

Take the square root of 100.

$$y=\frac{-2±10}{2\times 2}$$

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

$$y=\frac{-2±10}{4}$$

Now solve the equation y=\frac{-2±10}{4} when ± is plus. Add -2 to 10.

$$y=\frac{8}{4}$$

Divide 8 by 4.

$$y=2$$

Now solve the equation y=\frac{-2±10}{4} when ± is minus. Subtract 10 from -2.

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

Divide -12 by 4.

$$y=-3$$

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

$$x=2+1$$

Add 1\times 2 to 1.

$$x=3$$

Now substitute -3 for y in the equation x=y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-3+1$$

Add -3 to 1.

$$x=-2$$

The system is now solved.

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