Solve {l}{x^2+y^2=10}{x=5y}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 10 } \\ { x = 5 y } \end{array} \right.$$

Solve for x, y

$$x=-\frac{5\sqrt{65}}{13}\approx -3.100868365\text{, }y=-\frac{\sqrt{65}}{13}\approx -0.620173673$$

$$x=\frac{5\sqrt{65}}{13}\approx 3.100868365\text{, }y=\frac{\sqrt{65}}{13}\approx 0.620173673$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract 5y from both sides.

$$x-5y=0$$

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.

$$x-5y=0,y^{2}+x^{2}=10$$

Solve x-5y=0 for x by isolating x on the left hand side of the equal sign.

$$x-5y=0$$

Subtract -5y from both sides of the equation.

$$x=5y$$

Substitute 5y for x in the other equation, y^{2}+x^{2}=10.

$$y^{2}+\left(5y\right)^{2}=10$$

Square 5y.

$$y^{2}+25y^{2}=10$$

Add y^{2} to 25y^{2}.

$$26y^{2}=10$$

Subtract 10 from both sides of the equation.

$$26y^{2}-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 5^{2} for a, 1\times 0\times 2\times 5 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\times 26\left(-10\right)}}{2\times 26}$$

Square 1\times 0\times 2\times 5.

$$y=\frac{0±\sqrt{-4\times 26\left(-10\right)}}{2\times 26}$$

Multiply -4 times 1+1\times 5^{2}.

$$y=\frac{0±\sqrt{-104\left(-10\right)}}{2\times 26}$$

Multiply -104 times -10.

$$y=\frac{0±\sqrt{1040}}{2\times 26}$$

Take the square root of 1040.

$$y=\frac{0±4\sqrt{65}}{2\times 26}$$

Multiply 2 times 1+1\times 5^{2}.

$$y=\frac{0±4\sqrt{65}}{52}$$

Now solve the equation y=\frac{0±4\sqrt{65}}{52} when ± is plus.

$$y=\frac{\sqrt{65}}{13}$$

Now solve the equation y=\frac{0±4\sqrt{65}}{52} when ± is minus.

$$y=-\frac{\sqrt{65}}{13}$$

There are two solutions for y: \frac{\sqrt{65}}{13} and -\frac{\sqrt{65}}{13}. Substitute \frac{\sqrt{65}}{13} for y in the equation x=5y to find the corresponding solution for x that satisfies both equations.

$$x=5\times \frac{\sqrt{65}}{13}$$

Now substitute -\frac{\sqrt{65}}{13} for y in the equation x=5y and solve to find the corresponding solution for x that satisfies both equations.

$$x=5\left(-\frac{\sqrt{65}}{13}\right)$$

The system is now solved.

$$x=5\times \frac{\sqrt{65}}{13},y=\frac{\sqrt{65}}{13}\text{ or }x=5\left(-\frac{\sqrt{65}}{13}\right),y=-\frac{\sqrt{65}}{13}$$