define all solutions in integers to diophantine equations:
56x+72y=40
ok - I'll write my steps, my problem is that I end up with no solutions at all - maybe someone can find where I went wrong ... ?
SO:
72 = 56(1) + 16
56 = 16(3) + 8
16 = 8(2)
so gcd(56,72) = 8
next:
8 = 56 - (16)(3)
= 56 - 3(72-56)
= 56(4) + 72(-3)
40 = 56(20) + 72(-15)
so x0=20 & y0=-15
so:
x = 20 + 9t, for some int. t
y = -15 - 7t, for some int. t
and I get:
-(15/7) > t > -(20/9)
but there are no integers that work.
did I do something stupid in algebra or can there really be no solutions?
well - ok - what I would normally do (what we always do in class and on homework) is what I did above:
x = 20 + 9t, for some int. t
y = -15 - 7t, for some int. t
so you set 20 + 9t > 0
where t > -(20/9)
and then -15 - 7t > 0
where t < -(15/7)
and so you get
-(15/7) > t > -(20/9)
then (normally) from there we would take all the values of t in that restriction
(say for instance you had .5>t>2.5 - so t=1&2, and then you would plug those t values back into your x=blah and y=blah - and those are your solutions)
but with this problem - there are no integers in that restriction so that makes me wanna say no solutions ... ?
no, note that what you were solving for were the positive integer solutions. you wanted both x and y to be greater than zero. so you set the expressions for x and y to be greater than zero respectively, and then solve for the range that t has to be within. no such restriction was given here. we accept negative as well as positive solutions as far as this problem is concerned