Solve {l}{x^2+y^2=20}{x+3y=2}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 20 } \\ { x + 3 y = 2 } \end{array} \right.$$

Solve for x, y

$$x=\frac{22}{5}=4.4\text{, }y=-\frac{4}{5}=-0.8$$

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

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.

$$x+3y=2,y^{2}+x^{2}=20$$

Solve x+3y=2 for x by isolating x on the left hand side of the equal sign.

$$x+3y=2$$

Subtract 3y from both sides of the equation.

$$x=-3y+2$$

Substitute -3y+2 for x in the other equation, y^{2}+x^{2}=20.

$$y^{2}+\left(-3y+2\right)^{2}=20$$

Square -3y+2.

$$y^{2}+9y^{2}-12y+4=20$$

Add y^{2} to 9y^{2}.

$$10y^{2}-12y+4=20$$

Subtract 20 from both sides of the equation.

$$10y^{2}-12y-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-3\right)^{2} for a, 1\times 2\left(-3\right)\times 2 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 10\left(-16\right)}}{2\times 10}$$

Square 1\times 2\left(-3\right)\times 2.

$$y=\frac{-\left(-12\right)±\sqrt{144-4\times 10\left(-16\right)}}{2\times 10}$$

Multiply -4 times 1+1\left(-3\right)^{2}.

$$y=\frac{-\left(-12\right)±\sqrt{144-40\left(-16\right)}}{2\times 10}$$

Multiply -40 times -16.

$$y=\frac{-\left(-12\right)±\sqrt{144+640}}{2\times 10}$$

Add 144 to 640.

$$y=\frac{-\left(-12\right)±\sqrt{784}}{2\times 10}$$

Take the square root of 784.

$$y=\frac{-\left(-12\right)±28}{2\times 10}$$

The opposite of 1\times 2\left(-3\right)\times 2 is 12.

$$y=\frac{12±28}{2\times 10}$$

Multiply 2 times 1+1\left(-3\right)^{2}.

$$y=\frac{12±28}{20}$$

Now solve the equation y=\frac{12±28}{20} when ± is plus. Add 12 to 28.

$$y=\frac{40}{20}$$

Divide 40 by 20.

$$y=2$$

Now solve the equation y=\frac{12±28}{20} when ± is minus. Subtract 28 from 12.

$$y=-\frac{16}{20}$$

Reduce the fraction \frac{-16}{20} to lowest terms by extracting and canceling out 4.

$$y=-\frac{4}{5}$$

There are two solutions for y: 2 and -\frac{4}{5}. Substitute 2 for y in the equation x=-3y+2 to find the corresponding solution for x that satisfies both equations.

$$x=-3\times 2+2$$

Multiply -3 times 2.

$$x=-6+2$$

Add -3\times 2 to 2.

$$x=-4$$

Now substitute -\frac{4}{5} for y in the equation x=-3y+2 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-3\left(-\frac{4}{5}\right)+2$$

Multiply -3 times -\frac{4}{5}.

$$x=\frac{12}{5}+2$$

Add -3\left(-\frac{4}{5}\right) to 2.

$$x=\frac{22}{5}$$

The system is now solved.

$$x=-4,y=2\text{ or }x=\frac{22}{5},y=-\frac{4}{5}$$