Solve {l}{x+y=6}{x^2+y^2=20}

$$\left. \begin{array} { l } { x + y = 6 } \\ { x ^ { 2 } + y ^ { 2 } = 20 } \end{array} \right.$$

Solve for x, y

$$x=4\text{, }y=2$$

$$x=2\text{, }y=4$$

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.

$$x+y=6,y^{2}+x^{2}=20$$

Solve x+y=6 for x by isolating x on the left hand side of the equal sign.

$$x+y=6$$

Subtract y from both sides of the equation.

$$x=-y+6$$

Substitute -y+6 for x in the other equation, y^{2}+x^{2}=20.

$$y^{2}+\left(-y+6\right)^{2}=20$$

Square -y+6.

$$y^{2}+y^{2}-12y+36=20$$

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

$$2y^{2}-12y+36=20$$

Subtract 20 from both sides of the equation.

$$2y^{2}-12y+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 6\left(-1\right)\times 2 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 16}}{2\times 2}$$

Square 1\times 6\left(-1\right)\times 2.

$$y=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 16}}{2\times 2}$$

Multiply -4 times 1+1\left(-1\right)^{2}.

$$y=\frac{-\left(-12\right)±\sqrt{144-8\times 16}}{2\times 2}$$

Multiply -8 times 16.

$$y=\frac{-\left(-12\right)±\sqrt{144-128}}{2\times 2}$$

Add 144 to -128.

$$y=\frac{-\left(-12\right)±\sqrt{16}}{2\times 2}$$

Take the square root of 16.

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

The opposite of 1\times 6\left(-1\right)\times 2 is 12.

$$y=\frac{12±4}{2\times 2}$$

Multiply 2 times 1+1\left(-1\right)^{2}.

$$y=\frac{12±4}{4}$$

Now solve the equation y=\frac{12±4}{4} when ± is plus. Add 12 to 4.

$$y=\frac{16}{4}$$

Divide 16 by 4.

$$y=4$$

Now solve the equation y=\frac{12±4}{4} when ± is minus. Subtract 4 from 12.

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

Divide 8 by 4.

$$y=2$$

There are two solutions for y: 4 and 2. Substitute 4 for y in the equation x=-y+6 to find the corresponding solution for x that satisfies both equations.

$$x=-4+6$$

Add -4 to 6.

$$x=2$$

Now substitute 2 for y in the equation x=-y+6 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-2+6$$

Add -2 to 6.

$$x=4$$

The system is now solved.

$$x=2,y=4\text{ or }x=4,y=2$$