Diophantine equation

[mathlgs]

I need some help with my homework problem. The question is:

 

[TchrWill]

Re: Diophantine equation: 25x + 35y = 380 (Number Theory)

Find all solutions that are positive integers of the equation:

. . .25x + 35y = 380

1--25x + 35y = 380
2--Dividing through by the lowest coefficient yields
....1x + 1y + 10y/25 = 15 + 5/25
3--(10y - 5)/25 must be an integer
4--(10y - 5)/25 = (2y - 1)/5 (we want the coefficient of y to be 1)
5--(2y - 1)/5 is also an integer as is (6y - 3)/5
6--Dividing through by 5 yields 1y + y/5 - 3/5
7--(y - 3)/5 must bs an integer k making y = 5k + 3
8--Substituting back inti (1) yields 25x + 175k + 105 = 380 making
....x = 11-7k
9--k can only be 0 and 1 for positive integer answers.
....k......0......1
...x......11.....4
...y......3......8
10--checking:
......25(11) + 35(4) = 275 + 105 = 380
......25(4) += 35(8) = 100 + 280 = 380

 

[tkhunny]

Positive Integers

x > 0
y > 0

25x + 35 = 380 ==> x = 13.8

25 + 35y = 380 ==> y = 10.14

0 < x < 14
0 < y < 11

There are not very many to try, even with the "Brute Force" Method.

A little smarter.

25x + 35y = 380 ==> 5x + 7y = 76

x = (76 - 7y)/5 -- How many of those possibly y-values produce integer multiples of 5 from 76-7y? y = 3, 8