y = 2x+3

50x + y^2 = 2

asked by guest
on Sep 23, 2024 at 2:16 pm



You asked:

Solve the system of equations:\[\begin{aligned}y &= 2 x + 3\\50 x + {y}^{2} &= 2\end{aligned}\]

MathBot Answer:

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

\[x = - \frac{31}{4} - \frac{\sqrt{933}}{4}, y = - \frac{\sqrt{933}}{2} - \frac{25}{2}\]\[x = - \frac{31}{4} + \frac{\sqrt{933}}{4}, y = - \frac{25}{2} + \frac{\sqrt{933}}{2}\]


Solve \(y = 2 x + 3\) for \(x\). \[x = \frac{y}{2} - \frac{3}{2}\]Substitute \(\frac{y}{2} - \frac{3}{2}\) for \(x\) in \(50 x + y^{2} = 2\) and simplify. $$\begin{aligned}50 x + y^{2} &= 2 \\ 50 \left(\frac{y}{2} - \frac{3}{2}\right) + y^{2} &= 2 \\ y^{2} + 25 y &= 77 \\y^{2} + 25 y - 77 &= 0 \\ y &= \frac{-(25) \pm \sqrt{(25)^{2} - 4(1)(-77)}}{2(25)} \\ y = - \frac{25}{2} + \frac{\sqrt{933}}{2}&, y = - \frac{\sqrt{933}}{2} - \frac{25}{2}\end{aligned}$$Substitute \(- \frac{25}{2} + \frac{\sqrt{933}}{2}\) into \(y = 2 x + 3\) to solve for \(x\). \[\begin{aligned}- \frac{25}{2} + \frac{\sqrt{933}}{2} &= 2 x + 3\\- 2 x &= \frac{31}{2} - \frac{\sqrt{933}}{2}\\x &= - \frac{31}{4} + \frac{\sqrt{933}}{4}\end{aligned}\]This yields the following solution. $$\begin{aligned}x = - \frac{31}{4} + \frac{\sqrt{933}}{4},\,y = - \frac{25}{2} + \frac{\sqrt{933}}{2}\end{aligned}$$Substitute \(- \frac{\sqrt{933}}{2} - \frac{25}{2}\) into \(y = 2 x + 3\) to solve for \(x\). \[\begin{aligned}- \frac{\sqrt{933}}{2} - \frac{25}{2} &= 2 x + 3\\- 2 x &= \frac{\sqrt{933}}{2} + \frac{31}{2}\\x &= - \frac{31}{4} - \frac{\sqrt{933}}{4}\end{aligned}\]This yields the following solution. $$\begin{aligned}x = - \frac{31}{4} - \frac{\sqrt{933}}{4},\,y = - \frac{\sqrt{933}}{2} - \frac{25}{2}\end{aligned}$$