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Find all integer solutions of the equations: \begin{cases} 6x+21y&=&33 \\ 14x-49y&=&13 \end{cases}

I'm not sure how to find all integer solutions for a, but I know there are no integer solutions for b, because $\mbox{gcd}(14,49)=7$ and $7$ does not divide $13$.

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In a), first step is to divide through by 3. – Gerry Myerson Sep 18 '12 at 23:46
I know that. Which gives 2x+7y=11. By looking at it I can tell one integer solution is x=2 and y=1, but I don't know how to find all integer solutions. – Pink Panda Sep 18 '12 at 23:47
Good. Now, solve for $x$ and figure out what $y$ has to do to make $x$ an integer. – Gerry Myerson Sep 19 '12 at 0:05

For a, can you find values $a$ and $b$ so that $2(x+a)+7(y+b)=2x+7y$? If so, then given one solution (which you have) you can get more by $x'=x+na, y'=y+nb$ for any integer $n$. Then you need to prove that these are all the solutions.

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I'm not sure I follow, the only values for a, and b that would make 2(x+a)+7(y+b)=2x+7y would be a,b=0. What does this tell me? – Pink Panda Sep 19 '12 at 15:55
@PinkPanda: that is not true. You can cancel the terms in $x$ and $y$, leaving $2a+7b=0$. Doesn't that have a solution in the integers? In fact, many of them? – Ross Millikan Sep 19 '12 at 16:03
Oh, I see. It does have many, such as a=-7 and b=2. Still don't see how I would solve for all integer solutions. Do I solve for a and then solve for b? a=-7b/2 and b=-2a/7 – Pink Panda Sep 19 '12 at 16:17
@PinkPanda: so in addition to your solution of $(2,1)$, you can have $(2-7,1+2), (2-15\cdot 7, 1+15\cdot 2), \ldots $ – Ross Millikan Sep 19 '12 at 16:45

2x+7y=11 => 2x= 11-7y

11-7y must be even. For xy>0 your solution is unique

For xy<0, y must be an odd so that multiplied by 7 will make an odd and finally added with 11, make an even. So y can be every odd number, there will always be a x for that.

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