We have the following system of linear equations:

1. \( y = 4x + 7 \)
2. \( 2x + 5y = -31 \)

We can solve this system using substitution, as the first equation already expresses \( y \) in terms of \( x \).

### Steps:

1. Substitute \( y = 4x + 7 \) into the second equation \( 2x + 5y = -31 \):
   \[
   2x + 5(4x + 7) = -31
   \]

2. Distribute \( 5 \) on the left-hand side:
   \[
   2x + 20x + 35 = -31
   \]

3. Combine like terms:
   \[
   22x + 35 = -31
   \]

4. Subtract 35 from both sides:
   \[
   22x = -31 - 35
   \]
   \[
   22x = -66
   \]

5. Divide both sides by 22 to find \( x \):
   \[
   x = \frac{-66}{22} = -3
   \]

6. Now substitute \( x = -3 \) back into the first equation \( y = 4x + 7 \) to find \( y \):
   \[
   y = 4(-3) + 7 = -12 + 7 = -5
   \]

### Solution:
\[
x = -3, \quad y = -5
\]

Would you like more details or have any questions?

Here are 5 related questions you might consider:
1. How would you solve the system using the elimination method?
2. Can you graph both equations and identify their intersection point?
3. What happens if you substitute \( y = -5 \) directly into the second equation to check the solution?
4. How does the substitution method work for nonlinear equations?
5. How can you verify solutions of a system of equations?

**Tip:** When solving a system of equations, always substitute your solution back into both original equations to verify correctness.

Learn how to solve the system of linear equations y = 4x + 7 and 2x + 5y = -31 using the substitution method. This step-by-step solution covers important algebraic concepts and is suitable for students in Grades 8-10.

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