x^2+y^2=25

y=3/4(x)

asked by guest
on Sep 20, 2024 at 4:25 am



You asked:

Solve the system of equations:\[\begin{aligned}{x}^{2} + {y}^{2} &= 25\\y &= \frac{3}{4 x}\end{aligned}\]

MathBot Answer:

The system of equations has \(4\) solutions.

\[x = - \frac{4 \left(-5 + \sqrt{\frac{25}{2} - \frac{\sqrt{2491}}{4}}\right) \sqrt{\frac{25}{2} - \frac{\sqrt{2491}}{4}} \left(\sqrt{\frac{25}{2} - \frac{\sqrt{2491}}{4}} + 5\right)}{3}, y = \sqrt{\frac{25}{2} - \frac{\sqrt{2491}}{4}}\]\[x = - \frac{4 \left(-5 + \sqrt{\frac{\sqrt{2491}}{4} + \frac{25}{2}}\right) \sqrt{\frac{\sqrt{2491}}{4} + \frac{25}{2}} \left(\sqrt{\frac{\sqrt{2491}}{4} + \frac{25}{2}} + 5\right)}{3}, y = \sqrt{\frac{\sqrt{2491}}{4} + \frac{25}{2}}\]\[x = \frac{4 \left(-5 - \sqrt{\frac{25}{2} - \frac{\sqrt{2491}}{4}}\right) \left(5 - \sqrt{\frac{25}{2} - \frac{\sqrt{2491}}{4}}\right) \sqrt{\frac{25}{2} - \frac{\sqrt{2491}}{4}}}{3}, y = - \sqrt{\frac{25}{2} - \frac{\sqrt{2491}}{4}}\]\[x = \frac{4 \left(-5 - \sqrt{\frac{\sqrt{2491}}{4} + \frac{25}{2}}\right) \left(5 - \sqrt{\frac{\sqrt{2491}}{4} + \frac{25}{2}}\right) \sqrt{\frac{\sqrt{2491}}{4} + \frac{25}{2}}}{3}, y = - \sqrt{\frac{\sqrt{2491}}{4} + \frac{25}{2}}\]

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