Solve {c}{x^2+y^2=40}{y=3x}

$$\left. \begin{array} { c } { x ^ { 2 } + y ^ { 2 } = 40 } \\ { y = 3 x } \end{array} \right.$$

Solve for x, y

$$x=-2\text{, }y=-6$$

$$x=2\text{, }y=6$$

Steps Using Substitution and Quadratic Formula

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

$$y-3x=0$$

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.

$$y-3x=0,x^{2}+y^{2}=40$$

Solve y-3x=0 for y by isolating y on the left hand side of the equal sign.

$$y-3x=0$$

Subtract -3x from both sides of the equation.

$$y=3x$$

Substitute 3x for y in the other equation, x^{2}+y^{2}=40.

$$x^{2}+\left(3x\right)^{2}=40$$

Square 3x.

$$x^{2}+9x^{2}=40$$

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

$$10x^{2}=40$$

Subtract 40 from both sides of the equation.

$$10x^{2}-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 3^{2} for a, 1\times 0\times 2\times 3 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 10\left(-40\right)}}{2\times 10}$$

Square 1\times 0\times 2\times 3.

$$x=\frac{0±\sqrt{-4\times 10\left(-40\right)}}{2\times 10}$$

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

$$x=\frac{0±\sqrt{-40\left(-40\right)}}{2\times 10}$$

Multiply -40 times -40.

$$x=\frac{0±\sqrt{1600}}{2\times 10}$$

Take the square root of 1600.

$$x=\frac{0±40}{2\times 10}$$

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

$$x=\frac{0±40}{20}$$

Now solve the equation x=\frac{0±40}{20} when ± is plus. Divide 40 by 20.

$$x=2$$

Now solve the equation x=\frac{0±40}{20} when ± is minus. Divide -40 by 20.

$$x=-2$$

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

$$y=3\times 2$$

Multiply 3 times 2.

$$y=6$$

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

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

Multiply 3 times -2.

$$y=-6$$

The system is now solved.

$$y=6,x=2\text{ or }y=-6,x=-2$$