find the two numbers that are twice as far from 15 as they are from 24

24 + x =15 + 2x

x = 9

1st # 24 + 9 = 33

15 + 2x = 24 - x

3x = 24 - 15
3x = 9
x = 3

2nd # 15 + 6 = 21

Your answers are correct.

Ah, finding numbers can be quite the task, but fear not, dear human! Let's embark on this mathematical journey together and solve this puzzle.

Let's assume the first number is X, and the second number is Y.

According to the problem, X and Y are twice as far from 15 as they are from 24. Mathematically, we can write the following equation:

|X - 15| = 2 * |X - 24|

To make things a bit less complicated, let's analyze two separate cases:

Case 1: X - 15 = 2 * (X - 24)
Case 2: X - 15 = -2 * (X - 24)

Now, let's solve these equations, one at a time.

Case 1:
X - 15 = 2X - 48
48 - 15 = 2X - X
33 = X

Case 2:
X - 15 = -2X + 48
X + 2X = 48 + 15
3X = 63
X = 21

So, we have two possible values for X: 33 and 21.

Now, if we substitute those values back into the original equation, we can find the corresponding values for Y:

For X = 33:
|33 - 15| = 2 * |33 - 24|
18 = 2 * 9
18 = 18

For X = 21:
|21 - 15| = 2 * |21 - 24|
6 = 2 * 3
6 = 6

Therefore, the two numbers are either 33 and 18, or 21 and 6.

Voilà! We've found the numbers that satisfy the given condition.

To find the two numbers that are twice as far from 15 as they are from 24, let's break down the problem step by step:

1. Let's assume the two numbers we need to find are x and y.

2. The distance between 15 and x is twice the distance between x and 24. Mathematically, this can be represented as:
|15 - x| = 2 * |x - 24|

3. We need to solve this equation to find the values of x and y. Let's first solve it for x.

4. Solve the equation for x:
|15 - x| = 2 * |x - 24|

Since we have absolute values, we need to consider two cases:

Case 1: (15 - x) = 2 * (x - 24)
Case 2: (15 - x) = -2 * (x - 24)

Now we'll solve each of these cases separately.

Case 1:
15 - x = 2x - 48
3x = 63
x = 21

Case 2:
15 - x = -2x + 48
3x = 33
x = 11

Therefore, we have two potential values for x: 21 and 11.

5. Now that we have the possible values of x, we can find the corresponding values of y.

For x = 21:
The distance between 15 and 21 is (21 - 15) = 6. The distance between 21 and 24 is (24 - 21) = 3.
Since the distance between 15 and 21 is twice the distance between 21 and 24, this configuration is valid.

For x = 11:
The distance between 15 and 11 is (15 - 11) = 4. The distance between 11 and 24 is (24 - 11) = 13.
Since the distance between 15 and 11 is not twice the distance between 11 and 24, this configuration is not valid.

6. Therefore, the two numbers that are twice as far from 15 as they are from 24 are x = 21 and y = 15.