To determine the number of ordered pairs $(x, y)$ that satisfy the given system of equations:

1. $3x + 2y = 1$
2. $y = x^2 - 4x - 5$

we can substitute the expression for $y$ from the second equation into the first equation and solve for $x$.

Step-by-Step Solution:

1. Substitute $y$ from the second equation into the first equation:

$3x + 2(x^2 - 4x - 5) = 1$

2. Expand and simplify the equation:

$3x + 2x^2 - 8x - 10 = 1$

Combine like terms:

$2x^2 - 5x - 11 = 0$

3. Solve the quadratic equation for $x$:

To solve for $x$, we use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 2$, $b = -5$, and $c = -11$.

Calculate the discriminant $\Delta = b^2 - 4ac$:

$\Delta = (-5)^2 - 4 \cdot 2 \cdot (-11)$

$\Delta = 25 + 88$

$\Delta = 113$

Since the discriminant is positive ($113 > 0$), there are two distinct real solutions for $x$.

4. Find the values of $x$:

$x = \frac{-(-5) \pm \sqrt{113}}{2 \cdot 2}$

$x = \frac{5 \pm \sqrt{113}}{4}$

So, the two solutions for $x$ are:

$x_1 = \frac{5 + \sqrt{113}}{4}, \quad x_2 = \frac{5 - \sqrt{113}}{4}$

5. Find corresponding $y$ values for each $x$:

Substitute $x_1$ and $x_2$ back into the equation for $y$:

$y = x^2 - 4x - 5$

Calculate $y_1$ for $x_1$:

$y_1 = \left( \frac{5 + \sqrt{113}}{4} \right)^2 - 4 \left( \frac{5 + \sqrt{113}}{4} \right) - 5$

Calculate $y_2$ for $x_2$:

$y_2 = \left( \frac{5 - \sqrt{113}}{4} \right)^2 - 4 \left( \frac{5 - \sqrt{113}}{4} \right) - 5$

6. Conclusion:

There are two distinct ordered pairs $(x, y)$ that satisfy the system of equations.

---

Would you like more details or have any other questions? Here are some related questions:

1. How do you find the exact values of $y_1$ and $y_2$ in this context?
2. What is the geometric interpretation of the solution for this system of equations?
3. Can you graph both equations to visualize their intersection points?
4. What if the discriminant were zero or negative—how would that change the number of solutions?
5. How can this method be applied to solve other types of nonlinear systems?

Tip: Solving systems of equations by substitution is particularly useful when one equation is linear and the other is nonlinear.