The problem asks to solve the system of equations:

1. $y = 45x + 29$
2. $y = x^2 + 49x - 16$

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

Since both equations are equal to $y$, set them equal to each other:

$45x + 29 = x^2 + 49x - 16$

Step 2: Rearrange the equation

Reorganize all terms to one side of the equation:

$x^2 + 49x - 16 - 45x - 29 = 0$

Simplify:

$x^2 + 4x - 45 = 0$

Step 3: Solve the quadratic equation

The quadratic equation is:

$x^2 + 4x - 45 = 0$

We solve it using factoring, if possible. The factors of $-45$ that add to $4$ are $9$ and $-5$:

$(x + 9)(x - 5) = 0$

Thus, the solutions for $x$ are:

$x = -9 \quad \text{or} \quad x = 5$

Step 4: Solve for $y$

Substitute $x = -9$ and $x = 5$ into the first equation ($y = 45x + 29$) to find the corresponding $y$-values:

1. For $x = -9$: $y = 45(-9) + 29 = -405 + 29 = -376$

2. For $x = 5$: $y = 45(5) + 29 = 225 + 29 = 254$

Final Answer:

The solutions to the system are:

$(x, y) = (-9, -376) \quad \text{and} \quad (x, y) = (5, 254)$

Would you like a step-by-step verification of this solution or help with graphing?

Here are 5 follow-up questions:

1. How would you graphically represent these solutions on a coordinate plane?
2. What does it mean when two equations in a system share these points?
3. Can we solve this system using a different method, such as substitution or elimination?
4. What happens if the quadratic equation cannot be factored?
5. How do we determine whether the solutions are correct for both equations?

Tip: Always substitute your solutions back into the original equations to verify their correctness!