We are tasked with solving a problem related to a three-digit number $xyz$, where $x$, $y$, and $z$ represent the hundreds, tens, and ones digits respectively.

Key Information:

1. Sum of digits: $x + y + z = 12$.
2. Sum of hundreds and 3 times the ones digit is 5 times the tens digit: $x + 3z = 5y$.
3. New number after exchanging hundreds and ones is 96 less than 3 times the original: The new number formed by exchanging $x$ and $z$ becomes $100z + 10y + x$. This is related to the original number $100x + 10y + z$ through the equation: $100z + 10y + x = 3(100x + 10y + z) - 96$
4. We need to show that $97z = 299x + 20y - 96$.

Setting up the system of equations:

1. From the sum of the digits: x + y + z = 12 \tag{1}

2. From the second condition about the sum of digits: x + 3z = 5y \tag{2}

3. From the condition about swapping the digits: $100z + 10y + x = 3(100x + 10y + z) - 96$ Expanding the right-hand side: $100z + 10y + x = 300x + 30y + 3z - 96$ Simplifying this equation: $100z + 10y + x - 3z = 300x + 30y - 96$ $97z + 10y + x = 300x + 30y - 96$ Rearranging terms: 97z = 299x + 20y - 96 \tag{3}

Thus, we've shown that the equation holds. Now we can use these equations to solve for $x$, $y$, and $z$.

Would you like to proceed with solving the system of equations for the original three-digit number?

Follow-up Questions:

1. What are the steps to solve the system of equations $x + y + z = 12$, $x + 3z = 5y$, and $97z = 299x + 20y - 96$?
2. How does rearranging digits of a number affect its value mathematically?
3. Can we explore different methods to solve such systems of linear equations?
4. What is the significance of the condition involving the swap of digits in this problem?
5. Can we generalize this type of digit-swap problem to other numbers?

Tip:

Always substitute one equation into another to reduce the number of variables when solving systems of linear equations.