Hi Vickie,

Are you sure you wrote the problem correctly?

There is a whole number solution if the 3rd number is "a hundred more than the first number." If the third number is, in fact, a hundred times more than the third number, then your solutions will be decimals.

I will help you solve the problem **assuming the 3rd number is a hundred
more than the first.**

This problem has three unknown values: the 1st, 2nd, and 3rd number. To avoid using three different variables,
we can write each value in terms of the 1st number.

The 1st number is the value for which we are given no information. So
let's use x to represent that 1st number.

If we use to represent the 1st number, we can write the 2nd number as 5x (or five times x).

Then we can write the 3rd number as x+100 (or 100 more than the 1st number.)

We have:

**1st number = x**

**2nd number = 5x**

**3rd number = x +100**

The problem also tell us that the sum of the three number is 415.

To find the sum, we need to add all the values : **x + 5x + x + 100 = 415**

When we combine like terms our equation reads **7x + 100 = 415**

To solve for x, we need to isolate the x. That can be done by using inverse operations to "undo" what has been done to the x. 7x+100 means that x is being multiplied by 7, and is being added to 100. To undo,
subtract 100 from both sides. We are subtracting from both sides so that this remains an equality.

**7x + 100 - 100 = 415 - 100 **

**7x = 315**

To get x alone on the left side of the equation, we must divide both sides by 7.

**7x/7 = 315/7**

**x = 45**

Since x = 45, the 1st number is 45. That would make the 2nd number 225, since 5 times 45 is 225. Finally, the third number is 145, since 100 more than 45 is 145.

To check our work, we add all three values together to achieve a sum of 415

**45 + 225 + 145 = 315**

I hope this helped, Vickie!