x - y = 3

x^2+ y^2 = 13

asked by guest
on Nov 18, 2024 at 3:42 am



You asked:

Solve the system of equations:\[\begin{aligned}x - y &= 3\\{x}^{2} + {y}^{2} &= 13\end{aligned}\]

MathBot Answer:

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

\[x = \frac{3}{2} - \frac{\sqrt{17}}{2}, y = - \frac{\sqrt{17}}{2} - \frac{3}{2}\]\[x = \frac{3}{2} + \frac{\sqrt{17}}{2}, y = - \frac{3}{2} + \frac{\sqrt{17}}{2}\]

Solve \(x - y = 3\) for \(x\). \[x = y + 3\]Substitute \(y + 3\) for \(x\) in \(x^{2} + y^{2} = 13\) and simplify. $$\begin{aligned}x^{2} + y^{2} &amp= 13 \\ \left(y + 3\right)^{2} + y^{2} &= 13 \\ y^{2} + 3 y &= 2 \\y^{2} + 3 y - 2 &= 0 \\ y &= \frac{-(3) \pm \sqrt{(3)^{2} - 4(1)(-2)}}{2(3)} \\ y = - \frac{3}{2} + \frac{\sqrt{17}}{2}&, y = - \frac{\sqrt{17}}{2} - \frac{3}{2}\end{aligned}$$Substitute \(- \frac{3}{2} + \frac{\sqrt{17}}{2}\) into \(x - y = 3\) to solve for \(x\). \[\begin{aligned}x - \frac{\sqrt{17}}{2} + \frac{3}{2} &= 3\\x + \left(\frac{3}{2} - \frac{\sqrt{17}}{2}\right) &= 3\\x &= \frac{3}{2} + \frac{\sqrt{17}}{2}\end{aligned}\]This yields the following solution. $$\begin{aligned}x = \frac{3}{2} + \frac{\sqrt{17}}{2},\,y = - \frac{3}{2} + \frac{\sqrt{17}}{2}\end{aligned}$$Substitute \(- \frac{\sqrt{17}}{2} - \frac{3}{2}\) into \(x - y = 3\) to solve for \(x\). \[\begin{aligned}x + \frac{3}{2} + \frac{\sqrt{17}}{2} &= 3\\x + \left(\frac{3}{2} + \frac{\sqrt{17}}{2}\right) &= 3\\x &= \frac{3}{2} - \frac{\sqrt{17}}{2}\end{aligned}\]This yields the following solution. $$\begin{aligned}x = \frac{3}{2} - \frac{\sqrt{17}}{2},\,y = - \frac{\sqrt{17}}{2} - \frac{3}{2}\end{aligned}$$

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