Solve {l}{x^2+y^2=9}{y=25x}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 9 } \\ { y = 25 x } \end{array} \right.$$

Solve for x, y

$$x=-\frac{3\sqrt{626}}{626}\approx -0.119904115\text{, }y=-\frac{75\sqrt{626}}{626}\approx -2.997602876$$

$$x=\frac{3\sqrt{626}}{626}\approx 0.119904115\text{, }y=\frac{75\sqrt{626}}{626}\approx 2.997602876$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract 25x from both sides.

$$y-25x=0$$

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.

$$y-25x=0,x^{2}+y^{2}=9$$

Solve y-25x=0 for y by isolating y on the left hand side of the equal sign.

$$y-25x=0$$

Subtract -25x from both sides of the equation.

$$y=25x$$

Substitute 25x for y in the other equation, x^{2}+y^{2}=9.

$$x^{2}+\left(25x\right)^{2}=9$$

Square 25x.

$$x^{2}+625x^{2}=9$$

Add x^{2} to 625x^{2}.

$$626x^{2}=9$$

Subtract 9 from both sides of the equation.

$$626x^{2}-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 25^{2} for a, 1\times 0\times 2\times 25 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 626\left(-9\right)}}{2\times 626}$$

Square 1\times 0\times 2\times 25.

$$x=\frac{0±\sqrt{-4\times 626\left(-9\right)}}{2\times 626}$$

Multiply -4 times 1+1\times 25^{2}.

$$x=\frac{0±\sqrt{-2504\left(-9\right)}}{2\times 626}$$

Multiply -2504 times -9.

$$x=\frac{0±\sqrt{22536}}{2\times 626}$$

Take the square root of 22536.

$$x=\frac{0±6\sqrt{626}}{2\times 626}$$

Multiply 2 times 1+1\times 25^{2}.

$$x=\frac{0±6\sqrt{626}}{1252}$$

Now solve the equation x=\frac{0±6\sqrt{626}}{1252} when ± is plus.

$$x=\frac{3\sqrt{626}}{626}$$

Now solve the equation x=\frac{0±6\sqrt{626}}{1252} when ± is minus.

$$x=-\frac{3\sqrt{626}}{626}$$

There are two solutions for x: \frac{3\sqrt{626}}{626} and -\frac{3\sqrt{626}}{626}. Substitute \frac{3\sqrt{626}}{626} for x in the equation y=25x to find the corresponding solution for y that satisfies both equations.

$$y=25\times \frac{3\sqrt{626}}{626}$$

Now substitute -\frac{3\sqrt{626}}{626} for x in the equation y=25x and solve to find the corresponding solution for y that satisfies both equations.

$$y=25\left(-\frac{3\sqrt{626}}{626}\right)$$

The system is now solved.

$$y=25\times \frac{3\sqrt{626}}{626},x=\frac{3\sqrt{626}}{626}\text{ or }y=25\left(-\frac{3\sqrt{626}}{626}\right),x=-\frac{3\sqrt{626}}{626}$$