Questions: The sum of three numbers is 81 . The third number is 2 times the second. The second number is 9 less than the first. What are the numbers? First number: Second number: Third number:

Solution Steps

To solve this problem, we need to set up a system of equations based on the given relationships between the numbers. Let's denote the first number as \( x \), the second number as \( y \), and the third number as \( z \). According to the problem:

1. The sum of the three numbers is 81: \( x + y + z = 81 \)
2. The third number is 2 times the second: \( z = 2y \)
3. The second number is 9 less than the first: \( y = x - 9 \)

We can substitute the second and third equations into the first to solve for \( x \), \( y \), and \( z \).

We denote the first number as \( x \), the second number as \( y \), and the third number as \( z \). Based on the problem statement, we can establish the following equations:

1. \( x + y + z = 81 \)
2. \( z = 2y \)
3. \( y = x - 9 \)

We substitute equations (2) and (3) into equation (1): \[ x + (x - 9) + 2(x - 9) = 81 \] This simplifies to: \[ x + x - 9 + 2x - 18 = 81 \] Combining like terms gives: \[ 4x - 27 = 81 \]

Adding 27 to both sides: \[ 4x = 108 \] Dividing by 4: \[ x = 27 \]

Using \( x = 27 \) in equation (3): \[ y = 27 - 9 = 18 \] Using \( y = 18 \) in equation (2): \[ z = 2 \times 18 = 36 \]

Final Answer