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
$$7a + 4b=73$$
On
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.
On
We can rewrite the equations as \begin{align*} -11b+12c&=22-4a\\ 5b-4c&=17-a \end{align*} which gives us \begin{align*} b&=\frac{73-7a}4\\ c&=\frac{297-31a}{16} \end{align*}
We want find $a$'s such that the above expressions will be positive integers. From $297-31a>0$ we get $a<\frac{297}{31}$. For integer $a$'s this means that $a\le9$. (So we do not really have that many possibilities to try, only $a=1,2,\dots,9$.)
If we want $7a\equiv 73 \pmod 4$, this is equivalent to $a\equiv1\pmod4$.
Moreover, we want $31a\equiv297\pmod{16}$, which gives us $a\equiv7\pmod{16}$.
So $b$ and $c$ are integers only if $a=16k+7$.
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.