Daniel takes pleasure in experimenting with various kinds of toys. He bought three distinct types of toys for $25 in total. Furthermore, he possesses a toy that costs $20 more than three times the combined price of the other two toys. It's revealed that the total cost of the three toys multiplied by five, amounts to $125.
Identify the nature of the system (consistent, independent and number of solutions).

To solve this problem, let's define the variables as follows:
- Let $ x $ be the price of the first toy.
- Let $ y $ be the price of the second toy.
- Let $ z $ be the price of the third toy.

We are given the following information:
1. The total cost of the three toys is $25.
2. The cost of one toy is $20 more than three times the combined cost of the other two toys.
3. The total cost of the three toys multiplied by five equals $125.

Let's translate these into equations:

1. $ x + y + z = 25 $
2. One toy costs $20 more than three times the combined price of the other two toys. Let's assume this toy is $ z $, so:
   \[ z = 20 + 3(x + y) \]
3. The total cost of the three toys multiplied by five amounts to $125:
   \[ 5(x + y + z) = 125 \]

First, simplify the third equation:
\[ 5(x + y + z) = 125 \]
\[ x + y + z = \frac{125}{5} \]
\[ x + y + z = 25 \]

We notice that this equation is identical to the first one. Hence, the third condition does not provide any new information.

So, we are left with two equations:
\[ \begin{cases} 
x + y + z = 25 \\
z = 20 + 3(x + y) 
\end{cases} \]

To solve this system, we can substitute the first equation into the second:

From the first equation, we have:
\[ z = 25 - (x + y) \]

Substitute $ z $ in the second equation:
\[ 25 - (x + y) = 20 + 3(x + y) \]

Simplify and solve for $ x + y $:
\[ 25 - 20 = 4(x + y) \]
\[ 5 = 4(x + y) \]
\[ x + y = \frac{5}{4} \]

Now, substitute $ x + y = \frac{5}{4} $ back into the first equation:
\[ x + y + z = 25 \]
\[ \frac{5}{4} + z = 25 \]
\[ z = 25 - \frac{5}{4} \]
\[ z = \frac{100}{4} - \frac{5}{4} \]
\[ z = \frac{95}{4} \]

Since $ x + y = \frac{5}{4} $ and there are infinitely many pairs $(x, y)$ that satisfy this equation, we can conclude that the system has infinitely many solutions.

**Nature of the system:**
- **Consistent**: Because there is at least one solution.
- **Dependent**: Because there are infinitely many solutions.
- **Number of solutions**: Infinitely many solutions.

Do you want details or have any questions?

Here are 8 relative questions that expand the information:
1. What if the toy that costs $20 more than three times the combined price of the other two toys was $ x $ instead of $ z $?
2. How would the system change if the total cost of the three toys was $30 instead of $25?
3. Can you find specific values for $ x $ and $ y $ given additional constraints?
4. What happens if the combined price of $ x $ and $ y $ needs to be an integer?
5. How can you graphically represent this system of equations?
6. What if the cost of one toy was double the sum of the other two instead of triple?
7. How would you solve this system using matrix methods?
8. Can you formulate a

Learn how to solve a system of linear equations involving toy prices. Understand the concepts of linear equations and systems of equations with a detailed step-by-step solution. Suitable for students in Grades 7-9.

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