X+2Y=1

Y^2-10=2X

asked by guest
on Jan 21, 2025 at 8:47 pm



You asked:

Solve the system of equations:\[\begin{aligned}X + 2 Y &= 1\\{Y}^{2} - 10 &= 2 X\end{aligned}\]

MathBot Answer:

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

\[X = -3, Y = 2\]\[X = 13, Y = -6\]


Solve \(X + 2 Y = 1\) for \(X\). \[X = 1 - 2 Y\]Substitute \(1 - 2 Y\) for \(X\) in \(Y^{2} - 10 = 2 X\) and simplify. $$\begin{aligned}Y^{2} - 10 &= 2 X \\ Y^{2} - 10 &= 2 \left(1 - 2 Y\right) \\ Y^{2} + 4 Y &= 12 \\Y^{2} + 4 Y - 12 &= 0 \\ \left(Y - 2\right) \left(Y + 6\right) &= 0 \\ Y = -6&, Y = 2\end{aligned}$$Substitute \(-6\) into \(X + 2 Y = 1\) to solve for \(X\). \[\begin{aligned}X - 12 &= 1\\X &= 13\end{aligned}\]This yields the following solution. $$\begin{aligned}X = 13,\,Y = -6\end{aligned}$$Substitute \(2\) into \(X + 2 Y = 1\) to solve for \(X\). \[\begin{aligned}X + 4 &= 1\\X &= -3\end{aligned}\]This yields the following solution. $$\begin{aligned}X = -3,\,Y = 2\end{aligned}$$