Let's define the variables and then set up the system of linear equations based on the given problem:

• Let $x$ be the cost of the first toy.
• Let $y$ be the cost of the second toy.
• Let $z$ be the cost of the third toy.
• Let $w$ be the cost of the fourth toy.

Now, let's translate the given information into equations:

1. Total cost of the three toys:
   Daniel bought three toys for a total of $25.
   $x + y + z = 25$

2. Cost of the fourth toy:
   The fourth toy costs $20 more than three times the combined price of the other two toys (excluding $z$).
   $w = 3(x + y) + 20$

3. Product of total cost and 5:
   The problem states that the total cost of the three toys multiplied by five equals $125.
   $5(x + y + z) = 125$

Let's simplify this third equation:

$5(x + y + z) = 125$ $x + y + z = \frac{125}{5}$ $x + y + z = 25$

Interestingly, this equation is identical to the first equation, meaning we don't get any new information from it.

So, the system of equations that represents the situation is:

1) & \quad x + y + z = 25 \\ 2) & \quad w = 3(x + y) + 20 \\ \end{aligned}$$ These two equations describe the relationship between the costs of the toys. Do you want any further details or have any questions? Here are 8 relative questions for deeper understanding: 1. How can we find the specific values of $$x$$, $$y$$, and $$z$$ using these equations? 2. What does the equation $$w = 3(x + y) + 20$$ imply about the fourth toy's cost in relation to the others? 3. How could the problem change if the cost of the fourth toy was directly given? 4. What if the total cost of the three toys was different, say $30? How would that affect the system of equations? 5. Can the system of equations be solved using matrix methods? How? 6. How does the repetition of $$x + y + z = 25$$ affect the solution of the system? 7. What assumptions are we making about the cost of each toy? 8. How can you check the solution once you find the values of $$x$$, $$y$$, and $$z$$? **Tip:** Always look for overlapping or repeated information in a problem to avoid redundant equations, which might simplify the system.