SOLUTION: Sum of three numbers is twenty five the second number is twice the first the third number exceed the second by five find the numbers

Question 754317: Sum of three numbers is twenty five the second number is twice the first the third number exceed the second by five find the numbers

Answer by anteater(7) (Show Source): You can put this solution on YOUR website!
Call your first number x.

Your second number is twice the first, so it is 2x.

Your third number exceeds the second number by 5, so it is 2x + 5.

When you add these three numbers up, the sum is 25:

x + 2x + (2x + 5) = 25

Combine "like" terms:

5x + 5 = 25

Now, isolate (or solve for) x. First, subtract 5 from both sides of the equation:

5x + 5 - 5 = 25 - 5

5x = 20

Then, divide both sides of the equation by 5:

5x/5 = 20/5

x = 4

So our first number is 4.

The second number is twice this, or 8.

The third number exceeds 8 by 5. Since 8 + 5 = 13, 13 is our third number.
Finally,

4 + 8 + 13 = 25

I hope this was helpful! :)

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