Solve {l}{x^2+y^2=5}{x-y=3}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 5 } \\ { x - y = 3 } \end{array} \right.$$

Solve for x, y

$$x=2\text{, }y=-1$$

$$x=1\text{, }y=-2$$

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.

$$x-y=3,y^{2}+x^{2}=5$$

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}=5.

$$y^{2}+\left(y+3\right)^{2}=5$$

Square y+3.

$$y^{2}+y^{2}+6y+9=5$$

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

$$2y^{2}+6y+9=5$$

Subtract 5 from both sides of the equation.

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

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

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

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

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

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

$$y=\frac{-6±\sqrt{36-8\times 4}}{2\times 2}$$

Multiply -8 times 4.

$$y=\frac{-6±\sqrt{36-32}}{2\times 2}$$

Add 36 to -32.

$$y=\frac{-6±\sqrt{4}}{2\times 2}$$

Take the square root of 4.

$$y=\frac{-6±2}{2\times 2}$$

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

$$y=\frac{-6±2}{4}$$

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

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

Divide -4 by 4.

$$y=-1$$

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

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

Divide -8 by 4.

$$y=-2$$

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

$$x=-1+3$$

Add -1 to 3.

$$x=2$$

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

$$x=-2+3$$

Add -2 to 3.

$$x=1$$

The system is now solved.

$$x=2,y=-1\text{ or }x=1,y=-2$$