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

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

Solve for y, x

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

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

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract 3x from both sides.

$$y-3x=2$$

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.

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

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

$$y-3x=2$$

Subtract -3x from both sides of the equation.

$$y=3x+2$$

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

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

Square 3x+2.

$$x^{2}+9x^{2}+12x+4=20$$

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

$$10x^{2}+12x+4=20$$

Subtract 20 from both sides of the equation.

$$10x^{2}+12x-16=0$$

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

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

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

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

Multiply -4 times 1+1\times 3^{2}.

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

Multiply -40 times -16.

$$x=\frac{-12±\sqrt{144+640}}{2\times 10}$$

Add 144 to 640.

$$x=\frac{-12±\sqrt{784}}{2\times 10}$$

Take the square root of 784.

$$x=\frac{-12±28}{2\times 10}$$

Multiply 2 times 1+1\times 3^{2}.

$$x=\frac{-12±28}{20}$$

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

$$x=\frac{16}{20}$$

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

$$x=\frac{4}{5}$$

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

$$x=-\frac{40}{20}$$

Divide -40 by 20.

$$x=-2$$

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

$$y=3\times \frac{4}{5}+2$$

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

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

Add \frac{4}{5}\times 3 to 2.

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

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

$$y=3\left(-2\right)+2$$

Multiply 3 times -2.

$$y=-6+2$$

Add -2\times 3 to 2.

$$y=-4$$

The system is now solved.

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