x-2y=5

x^2-y^2-3xy=0

asked by guest
on Oct 06, 2024 at 1:25 am



You asked:

Solve the system of equations:\[\begin{aligned}x - 2 y &= 5\\{x}^{2} - {y}^{2} - 3 x y &= 0\end{aligned}\]

MathBot Answer:

The system of equations has \(2\) solutions.

\[x = \frac{20}{3} - \frac{5 \sqrt{13}}{3}, y = \frac{5}{6} - \frac{5 \sqrt{13}}{6}\]\[x = \frac{5 \sqrt{13}}{3} + \frac{20}{3}, y = \frac{5}{6} + \frac{5 \sqrt{13}}{6}\]

Solve \(x - 2 y = 5\) for \(x\). \[x = 2 y + 5\]Substitute \(2 y + 5\) for \(x\) in \(x^{2} - 3 x y - y^{2} = 0\) and simplify. $$\begin{aligned}x^{2} - 3 x y - y^{2} &= 0 \\ \left(2 y + 5\right)^{2} - 3 \left(2 y + 5\right) y - y^{2} &= 0 \\ 3 y^{2} - 5 y &= 25 \\3 y^{2} - 5 y - 25 &= 0 \\ y &= \frac{-(-5) \pm \sqrt{(-5)^{2} - 4(3)(-25)}}{2(-5)} \\ y = \frac{5}{6} - \frac{5 \sqrt{13}}{6}&, y = \frac{5}{6} + \frac{5 \sqrt{13}}{6}\end{aligned}$$Substitute \(\frac{5}{6} - \frac{5 \sqrt{13}}{6}\) into \(x - 2 y = 5\) to solve for \(x\). \[\begin{aligned}x - \frac{5}{3} + \frac{5 \cdot \sqrt{13}}{3} &= 5\\x - \left(\frac{5}{3} - \frac{5 \sqrt{13}}{3}\right) &= 5\\x &= \frac{20}{3} - \frac{5 \sqrt{13}}{3}\end{aligned}\]This yields the following solution. $$\begin{aligned}x = \frac{20}{3} - \frac{5 \sqrt{13}}{3},\,y = \frac{5}{6} - \frac{5 \sqrt{13}}{6}\end{aligned}$$Substitute \(\frac{5}{6} + \frac{5 \sqrt{13}}{6}\) into \(x - 2 y = 5\) to solve for \(x\). \[\begin{aligned}x - \frac{5 \cdot \sqrt{13}}{3} - \frac{5}{3} &= 5\\x - \left(\frac{5}{3} + \frac{5 \sqrt{13}}{3}\right) &= 5\\x &= \frac{5 \sqrt{13}}{3} + \frac{20}{3}\end{aligned}\]This yields the following solution. $$\begin{aligned}x = \frac{5 \sqrt{13}}{3} + \frac{20}{3},\,y = \frac{5}{6} + \frac{5 \sqrt{13}}{6}\end{aligned}$$

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