SOLUTION: Find all solutions (x,y) in positive integers to: 13x + 14y = 2008 I know we are supposed to use the Diophantine equation but I'm not comletely sure how to yet.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find all solutions (x,y) in positive integers to: 13x + 14y = 2008 I know we are supposed to use the Diophantine equation but I'm not comletely sure how to yet.       Log On


   

Question 618734: Find all solutions (x,y) in positive integers to:
13x + 14y = 2008
I know we are supposed to use the Diophantine equation but I'm not
comletely sure how to yet.
Found 2 solutions by ewatrrr, fcabanski:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Find all solutions (x,y) in positive integers to:

13x + 14y = 2008

x = xo + pt

y = yo - qt

(134,19) one solution

x = 134 + pt

y = 19 - qt

As GCD of 13 and 14 is 1, therefore:

x = 134 + 14t

y =  19 - 13t

Answer by fcabanski(1391) About Me  (Show Source):
You can put this solution on YOUR website!
Step one - use trial and error to find one solution for x,y.


In this case, use x=162 and y=-7: (13*162+14(-7) = 2106 - 98 = 2008)


The general solution is x=x(0) + pt and y = y(0) - qt where x(0) and y(0) are the values found from trial and error, p = b (coefficient of y) / GCD (a,b) and q = a (coefficient of x) / GCD(a,b) and t is an integer.


The GCD of a and b is 1. So p = 14/1 = 14 and q = 13/1 = 13.


So the solution in this case is x = 162+14t and y = -7 - 13t

Hope the solution helped. Sometimes you need more than a solution. Contact fcabanski@hotmail.com for online, private tutoring, or personalized problem solving (quick for groups of problems.)