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

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

Solve for x, y

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

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

Steps Using Substitution and Quadratic Formula

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

$$x-3y=0$$

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.

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

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

$$x-3y=0$$

Subtract -3y from both sides of the equation.

$$x=3y$$

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

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

Square 3y.

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

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

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

Subtract 40 from both sides of the equation.

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

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}.

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

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

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

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

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

Multiply -40 times -40.

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

Take the square root of 1600.

$$y=\frac{0±40}{2\times 10}$$

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

$$y=\frac{0±40}{20}$$

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

$$y=2$$

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

$$y=-2$$

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

$$x=3\times 2$$

Multiply 3 times 2.

$$x=6$$

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

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

Multiply 3 times -2.

$$x=-6$$

The system is now solved.

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