Solve {c}{x-y=3}{x^2+y^2=29}

$$\left\{ \begin{array} { c } { x - y = 3 } \\ { x ^ { 2 } + y ^ { 2 } = 29 } \end{array} \right.$$

Solve for x, y

$$x=5\text{, }y=2$$

$$x=-2\text{, }y=-5$$

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.

$$x-y=3,y^{2}+x^{2}=29$$

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}=29.

$$y^{2}+\left(y+3\right)^{2}=29$$

Square y+3.

$$y^{2}+y^{2}+6y+9=29$$

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

$$2y^{2}+6y+9=29$$

Subtract 29 from both sides of the equation.

$$2y^{2}+6y-20=0$$

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 -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-6±\sqrt{6^{2}-4\times 2\left(-20\right)}}{2\times 2}$$

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

$$y=\frac{-6±\sqrt{36-4\times 2\left(-20\right)}}{2\times 2}$$

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

$$y=\frac{-6±\sqrt{36-8\left(-20\right)}}{2\times 2}$$

Multiply -8 times -20.

$$y=\frac{-6±\sqrt{36+160}}{2\times 2}$$

Add 36 to 160.

$$y=\frac{-6±\sqrt{196}}{2\times 2}$$

Take the square root of 196.

$$y=\frac{-6±14}{2\times 2}$$

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

$$y=\frac{-6±14}{4}$$

Now solve the equation y=\frac{-6±14}{4} when ± is plus. Add -6 to 14.

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

Divide 8 by 4.

$$y=2$$

Now solve the equation y=\frac{-6±14}{4} when ± is minus. Subtract 14 from -6.

$$y=-\frac{20}{4}$$

Divide -20 by 4.

$$y=-5$$

There are two solutions for y: 2 and -5. Substitute 2 for y in the equation x=y+3 to find the corresponding solution for x that satisfies both equations.

$$x=2+3$$

Add 1\times 2 to 3.

$$x=5$$

Now substitute -5 for y in the equation x=y+3 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-5+3$$

Add -5 to 3.

$$x=-2$$

The system is now solved.

$$x=5,y=2\text{ or }x=-2,y=-5$$