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Solve the systems if $a$, $b$, and $c$ are positive integers:

$$4a-11b+12c=22$$

$$a+5b-4c=17$$

The answer is given in the back as $(7,6,5)$ but how can you approach three variables with only two equations?

I tried multiplying the second equation by 3 and adding I got this

$$7a + 4b=73$$

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    $\begingroup$ Welcome to stackexchange. You are more likely to get answers rather than downvotes and votes to close if you edit the question to show what you have tried and where you are stuck. $\endgroup$ – Ethan Bolker Jan 21 '18 at 20:56
  • $\begingroup$ I'm stuck at finding a way to get another equation. $\endgroup$ – skillpatrol Jan 21 '18 at 20:59
  • $\begingroup$ The solution for this system of equations will look something like this: $a=A_1+A_2*p$, $b=B_1+B_2*p$ and $c=p$. And from this, you should search integer solutions. $\endgroup$ – Botond Jan 21 '18 at 20:59
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    $\begingroup$ $7a\color{red}{+}4b=73$.& this has solutions in positive integers $(a,b)=(3,13)$ or $(7,6)$ ... so ... $\endgroup$ – Donald Splutterwit Jan 21 '18 at 21:10
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    $\begingroup$ Modulo refers to taking the remainder on division, so taking an equation modulo $4$ results in one of $0,1,2,$ or $3$. For integer equations you can often eliminate cases that way. I think $\bmod 7$ is more useful here. $\endgroup$ – Ross Millikan Jan 21 '18 at 21:30
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Given $7a+4b=73$ you can take it $\bmod 7$ to get $4b \equiv 3 \bmod 7, b \equiv 6 \bmod 7$. This gives $b=6$ or $13$ because $20$ is too large. Plugging in, you find $13$ is too large as well.

If you want to do it without the modulo operation, note that $a \lt 11$ because otherwise $7a \gt 73$, so you only have ten choices and can try them all.

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  • $\begingroup$ Ok, that narrows it down a lot. Thanks! $\endgroup$ – skillpatrol Jan 21 '18 at 21:37
  • $\begingroup$ Also note that $a$ must be odd, so that cuts the choices down to $5$. $\endgroup$ – B. Goddard Jan 21 '18 at 21:38
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Once you find one solution in integers, any other solution happens by adding an integer multiple of the coefficient cross product, namely $$ \langle -16, 28, 31 \rangle $$ You know the solution $ \langle 7,6,5 \rangle \; . $ Any other integer solution is $$ \langle 7-16t, 6+28t, 5+31t \rangle \; . $$ If $t > 0$ we get $7 - 16 t < 0.$ If $t < 0$ then $6 + 28 t < 0.$ It follows that $t=0,$ the only solution in positive integers is the given one. There are infinitely many integer solutions, they lie on the line I described, but that line passes only briefly through the first (positive) octant.

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  • $\begingroup$ Coefficient "cross product" is not in our textbook. $\endgroup$ – skillpatrol Jan 21 '18 at 21:42
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    $\begingroup$ @user685252 you have a difficult life. I wish you the strength to carry on. $\endgroup$ – Will Jagy Jan 21 '18 at 21:43

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