SOLUTION: find two consecutive integers such that the sum of one third of the first integer and one fourth of the second integer is 9.

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Question 898326: find two consecutive integers such that the sum of one third of the first integer and one fourth of the second integer is 9.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
THE ALGEBRA WAY:
n= the smaller of the two consecutive integers
n%2B1= the largerer of the two consecutive integers
The problem says that
n%2F3%2B%28n%2B1%29%2F4=9
Multiplying both sides of the equalsign times 3%2A4=12 we get
12%28n%2F3%2B%28n%2B1%29%2F4%29=12%2A9
12%2An%2F3%2B12%28n%2B1%29%2F4=108
4n%2B3%28n%2B1%29=108
4n%2B3n%2B3=108
7n%2B3=108
7n=108-3
7n=105
n=105%2F7
highlight%28n=15%29
The two consecutive integers are highlight%2815%29 and highlight%2816%29 .

ANOTHER WAY (Guess and check):
One third of the first integer must be an integer (a whole number),
and one fourth of the second integer must be an integer too.
Otherwise, one or both would be fractions that would add up to a fraction.
The first integer could be 0, 3, 6, 9, 12, 15, 18, 21, 24, ....,
and one third of that would be 0, 1, 2, 3, 4, 5, 6, 7, 8, ....
If we add 1 to each of those choices, the second integer would be 1, 4, 7, 13, 16, 19, 22, 25, ...., but not all of those numbers divide by 4 evenly.
Only 4, and 16 do.
If the second integer were 4 and the first one were 3,
one third of the first integer and one fourth of the second integer would be
3%2F3%2B4%2F4=1%2B1=2 , so 3 and 4 is not the answer.
If the second integer were 16 and the first one were 15,
one third of the first integer and one fourth of the second integer would be
15%2F3%2B16%2F4=5%2B4=9 , so highlight%2815%29 and highlight%2816%29 is the answer.