3x-2y+7=0, $x^{2}$ + $y^{2}$ =4x+9

asked by guest
on Dec 04, 2024 at 1:06 am



You asked:

Solve the system of equations:\[\begin{aligned}3 x - 2 y + 7 &= 0\\{x}^{2} + {y}^{2} &= 4 x + 9\end{aligned}\]

MathBot Answer:

The system of equations has one solution.

\[x = -1, y = 2\]


Solve \(3 x - 2 y + 7 = 0\) for \(x\). \[x = \frac{2 y}{3} - \frac{7}{3}\]Substitute \(\frac{2 y}{3} - \frac{7}{3}\) for \(x\) in \(x^{2} + y^{2} = 4 x + 9\) and simplify. $$\begin{aligned}x^{2} + y^{2} &= 4 x + 9 \\ \left(\frac{2 y}{3} - \frac{7}{3}\right)^{2} + y^{2} &= 4 \left(\frac{2 y}{3} - \frac{7}{3}\right) + 9 \\ y^{2} - 4 y &= -4 \\y^{2} - 4 y + 4 &= 0 \\ \left(y - 2\right)^{2} &= 0 \\ y = 2\end{aligned}$$Substitute \(2\) into \(3 x - 2 y + 7 = 0\) to solve for \(x\). \[\begin{aligned}3 x + 3 &= 0\\3 x &= -3\\x &= -1\end{aligned}\]This yields the following solution. $$\begin{aligned}x = -1,\,y = 2\end{aligned}$$