Let's solve this problem step by step by setting up equations based on the information provided.

Step 1: Define Variables

Let the costs of the three distinct types of toys be $x$, $y$, and $z$.

• $x$: Cost of the first toy
• $y$: Cost of the second toy
• $z$: Cost of the third toy

Step 2: Set Up the Equations

Based on the problem, we have the following information:

1. Total cost of the three toys: $x + y + z = 25$

2. The cost of the third toy: $z$ is $20 more than three times the combined price of the other two toys: $z = 20 + 3(x + y)$

3. Total cost multiplied by five: The total cost of the three toys multiplied by five is $125: $5(x + y + z) = 125$ Simplifying this equation: $x + y + z = \frac{125}{5} = 25$ This is consistent with our first equation.

Step 3: Substitute and Solve

We now have two key equations:

1. $x + y + z = 25$
2. $z = 20 + 3(x + y)$

Let's substitute the expression for $z$ from the second equation into the first equation:

$x + y + (20 + 3(x + y)) = 25$

Simplify the equation: $x + y + 20 + 3x + 3y = 25$ $4x + 4y + 20 = 25$ $4x + 4y = 5$

Step 4: Simplify Further

Divide the equation by 4: $x + y = \frac{5}{4} = 1.25$

Now, substitute $x + y = 1.25$ back into the equation for $z$: $z = 20 + 3(1.25)$ $z = 20 + 3.75 = 23.75$

Step 5: Conclusion

So, the costs of the toys are:

• $x + y = 1.25$
• $z = 23.75$

Since $x + y = 1.25$, there are infinitely many possibilities for $x$ and $y$ as long as their sum equals 1.25. For example:

• If $x = 0.25$, then $y = 1.00$
• If $x = 0.75$, then $y = 0.50$

Thus, the costs of the toys could be $0.25$, $1.00$, and $23.75$ or any other combination of $x$ and $y$ that satisfies $x + y = 1.25$.

Summary:

• $x$ and $y$ are any values that sum to $1.25$.
• $z = 23.75$.

Would you like more details or have any questions?

Here are some related questions:

1. What if the sum of $x + y$ were different?
2. How would the solution change if $z$ were only $15 more than twice the sum of $x$ and $y$?
3. Can you find a specific solution for $x$ and $y$ if $x = 0.5$?
4. What happens if $x + y + z$ had to be $30 instead of $25?
5. How would the equations change if $z$ were dependent on the difference rather than the sum of $x$ and $y$?
6. What if there were four toys instead of three?
7. Could the solution be represented graphically? How?
8. How would rounding affect the solutions for $x$ and $y$?

Tip: When dealing with equations involving sums, always check if the equation can be simplified early on to reduce complexity.