Question 1193931
.
Find three consecutive whole numbers such that twice the sum of the two smallest numbers 
is 10 more than three times the largest number.
~~~~~~~~~~~~~~


<pre>
Let the numbers be n, (n+1) and (n+2).


Write equation as you read the problem

    2*(n + (n+1)) = 3*(n+2) + 10.


Now simplify and find "n"

    2n + 2n + 2 = 3n + 6 + 10

    4n + 2 = 3n + 16

    4n - 3n = 16 - 2

       n    =    14


<U>ANSWER</U>.  The numbers are 14, 15 and 16.
</pre>

Solved, answered and explained.