Solve this system equation::

-2y^2+ 25x - y- 47 = 0

x - y + 1 = 0

asked by guest
on Jan 27, 2025 at 5:25 am



You asked:

Solve the system of equations:\[\begin{aligned}-\left( 2 {y}^{2} \right) + 25 x - y - 47 &= 0\\x - y + 1 &= 0\end{aligned}\]

MathBot Answer:

The system of equations has one solution.

\[x = 5, y = 6\]

Solve \(25 x - 2 y^{2} - y - 47 = 0\) for \(x\). \[x = \frac{2 y^{2}}{25} + \frac{y}{25} + \frac{47}{25}\]Substitute \(\frac{2 y^{2}}{25} + \frac{y}{25} + \frac{47}{25}\) for \(x\) in \(x - y + 1 = 0\) and simplify. $$\begin{aligned}x - y + 1 &= 0 \\ \left(\frac{2 y^{2}}{25} + \frac{y}{25} + \frac{47}{25}\right) - y + 1 &= 0 \\ y^{2} - 12 y &= -36 \\y^{2} - 12 y + 36 &= 0 \\ \left(y - 6\right)^{2} &= 0 \\ y = 6\end{aligned}$$Substitute \(6\) into \(25 x - 2 y^{2} - y - 47 = 0\) to solve for \(x\). \[\begin{aligned}25 x - 125 &= 0\\25 x &= 125\\x &= 5\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 5,\,y = 6\end{aligned}$$

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