x*y +1.5 y2y^{2} = 11.5

x+2y 3.25\sqrt{3.25} = 6.86

asked by guest
on Mar 26, 2025 at 6:20 pm



You asked:

Solve the system of equations:xy+1.5y2=11.5x+2y3.25=6.86\begin{aligned}x y + 1.5 {y}^{2} &= 11.5\\x + 2 y \sqrt{3.25} &= 6.86\end{aligned}

MathBot Answer:

The system of equations has 22 solutions.

x=34350−13⋅(10292150+343131075−i240911+3533212132150),y=10292150+343131075−i240911+3533212132150x = \frac{343}{50} - \sqrt{13} \cdot \left(\frac{1029}{2150} + \frac{343 \sqrt{13}}{1075} - \frac{i \sqrt{240911 + 3533212 \sqrt{13}}}{2150}\right), y = \frac{1029}{2150} + \frac{343 \sqrt{13}}{1075} - \frac{i \sqrt{240911 + 3533212 \sqrt{13}}}{2150}x=34350−13⋅(10292150+343131075+i240911+3533212132150),y=10292150+343131075+i240911+3533212132150x = \frac{343}{50} - \sqrt{13} \cdot \left(\frac{1029}{2150} + \frac{343 \sqrt{13}}{1075} + \frac{i \sqrt{240911 + 3533212 \sqrt{13}}}{2150}\right), y = \frac{1029}{2150} + \frac{343 \sqrt{13}}{1075} + \frac{i \sqrt{240911 + 3533212 \sqrt{13}}}{2150}

ii is the imaginary unit, defined as i2=−1i^2 = -1.
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