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The problem you have provided involves finding the costs of pencils, pens, and erasers from given linear equations.

Here’s the breakdown of the equations from the image:

1. \( 5 \text{ pencils} + 9 \text{ pens} + 13 \text{ erasers} = 190 \)
2. \( 9 \text{ pencils} + 5 \text{ pens} + 19 \text{ erasers} = 190 \)
3. \( 30 \text{ pencils} + 26 \text{ pens} + 57 \text{ erasers} = 380 \)

We need to solve this system of linear equations to find the cost of one pencil, one pen, and one eraser.

Let's denote the costs as:
- \( x \) for one pencil,
- \( y \) for one pen,
- \( z \) for one eraser.

Thus, the system becomes:
1. \( 5x + 9y + 13z = 190 \)
2. \( 9x + 5y + 19z = 190 \)
3. \( 30x + 26y + 57z = 380 \)

I'll solve this system of equations to find the values of \( x \), \( y \), and \( z \).The solution to the system of equations is:

- The cost of one pencil \( x = -\frac{817}{14} \approx -58.36 \)
- The cost of one pen \( y = -\frac{19}{14} \approx -1.36 \)
- The cost of one eraser \( z = 38 \)

However, these negative values for the cost of pencils and pens indicate that there might be an issue with the provided equations or the interpretation. Please double-check the problem or equations to ensure they are correct.

Would you like further clarification, or shall we explore possible ways to adjust the system?

Here are 5 related questions to help expand your understanding:
1. What happens if we eliminate one variable using substitution?
2. How can we interpret negative values in practical terms?
3. Could we explore alternative solutions using approximate values for the costs?
4. What are some common mistakes made when solving simultaneous equations?
5. How do you use matrix methods like Cramer's rule to solve similar problems?

**Tip:** When solving simultaneous equations, always verify the accuracy of the input equations to avoid unexpected results!

Find the cost of pencils, pens, and erasers from the given linear equations: 
1. 5 pencils + 9 pens + 13 erasers = 190 
2. 9 pencils + 5 pens + 19 erasers = 190 
3. 30 pencils + 26 pens + 57 erasers = 380

Learn how to solve a system of linear equations to find the cost of pencils, pens, and erasers. This problem involves solving three equations with three variables using algebraic methods. Suitable for students in Grades 10-12.

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