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.