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The system of equations given is:

\[
\begin{cases} 
x - 1 = (y + 3)^2 \\
4x - y = 40 
\end{cases}
\]

To solve this system, we can proceed with the following steps:

1. **Rewrite the first equation** to express \( x \) in terms of \( y \):
   \[
   x = (y + 3)^2 + 1
   \]

2. **Substitute this expression for \( x \) in the second equation**:
   \[
   4((y + 3)^2 + 1) - y = 40
   \]

3. **Expand and simplify**:
   \[
   4(y + 3)^2 + 4 - y = 40
   \]
   Expanding \( (y + 3)^2 \):
   \[
   4(y^2 + 6y + 9) + 4 - y = 40
   \]
   \[
   4y^2 + 24y + 36 + 4 - y = 40
   \]
   \[
   4y^2 + 23y = 0
   \]

4. **Factor the quadratic equation**:
   \[
   y(4y + 23) = 0
   \]
   So, \( y = 0 \) or \( y = -\frac{23}{4} \).

5. **Substitute each value of \( y \) back into the expression for \( x \)** to find the corresponding values of \( x \).

   - **For \( y = 0 \)**:
     \[
     x = (0 + 3)^2 + 1 = 9 + 1 = 10
     \]
     So, one solution is \( (x, y) = (10, 0) \).

   - **For \( y = -\frac{23}{4} \)**:
     \[
     x = \left(-\frac{23}{4} + 3\right)^2 + 1
     \]
     Simplifying inside the parentheses:
     \[
     -\frac{23}{4} + \frac{12}{4} = -\frac{11}{4}
     \]
     Then:
     \[
     x = \left(-\frac{11}{4}\right)^2 + 1 = \frac{121}{16} + 1 = \frac{121}{16} + \frac{16}{16} = \frac{137}{16}
     \]
     So, the other solution is \( \left(x, y\right) = \left(\frac{137}{16}, -\frac{23}{4}\right) \).

### Final Solutions:
The solutions to the system are:
\[
(x, y) = (10, 0) \quad \text{and} \quad \left(x, y\right) = \left(\frac{137}{16}, -\frac{23}{4}\right)
\]

Would you like a deeper breakdown of any part? 

Here are some related questions you might find helpful:

1. How can this type of system be solved using substitution?
2. What does each equation represent graphically?
3. How does completing the square help in solving quadratic equations?
4. Are there alternative methods to solve this system?
5. How does factoring simplify solving polynomial equations?

**Tip:** When solving systems of equations, look for ways to express one variable in terms of the other to simplify substitution.

Solve the system of equations: x - 1 = (y + 3)^2 and 4x - y = 40.

This tutorial explains how to solve the system of equations x - 1 = (y + 3)^2 and 4x - y = 40 using substitution. Covers key algebraic concepts such as quadratic equations, substitution, and factoring, making it suitable for high school students in Grades 10-12.

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