Show $\mathrm{gcd}(7a+5,4a+3)=1$. I have been trying to do this problem for a couple of days for better or worse.
I suppose that $d = \mathrm{gcd}(7a+5,4a+3)$. Since $4a+3=2(2a+1)+1$ it must be that $d$ is odd. I know that $d|(7a+5)$ and $d|(4a+3)$ so $d|(11a+8)$. I also know that $\mathrm{gcd}(a,b) \leq \mathrm{gcd}(a+b,a-b)$ but I haven't been able to get anything useful out of that yet.
I have just been trying to find some kind of contradiction assuming that $d>1$ but I haven't found anything definite.
 A: Do you know about the Bezout identity?
$$7(4a+3)-4(7a+5)=1$$
The choice of $7$ and $4$ as coefficients is guided by the fact that you need the $a$'s to cancel exactly.
A: The Euclidean algorithm, and ideas based on it, are frequently the best method for common divisor problems.
In this situation, we have a conundrum: we can't tell which number is larger and how many times the smaller one goes into the larger one! So, we have to decide on some other approach.
We could treat $a$ as being "big" so that we always try to cancel it out:
$$\begin{align} \gcd(4a+3, 7a+5)
&= \gcd(4a+3, 3a+2)
\\&= \gcd(a+1, 3a+2)
\\&= \gcd(a+1, a)
\\&= \gcd(1, a)
\\&= 1
\end{align}$$
Things turned out nicely; we might have been left with something like $\gcd(3,a)$; even if that doesn't give us the answer, it is still a much simpler expression than the origina.
We could have treated $a$ as "small" instead, so that each step is designed to reduce the constant term, rather than the coefficient of $a$. Unfortunately, this problem doesn't make for a good example of the difference between the approaches since it results in exactly the same steps.

And as an aside, don't be intimidated by variables! They can make things a little bit more complicated, but $a$ is just as much of a number as $7$ is, and much of what you learn to deal with numbers works equally well for decimals as they do for variables, and a lot of the rest still works just to a lesser and more complicated extent.
A: Hint $\rm\,\ d\mid 7a\!+\!5,4a\!+\!3\:$ becomes a linear system of equations modulo $\rm\,d,$ with determinant $= 1.\,$ But Cramer's rule works over any commutative ring when the determinant $=1\,$ (or invertible). 
$$\rm mod\ d\!:\ \bigg\lbrace\begin{array}{c} 7a+ 5\equiv 0 \\ \rm 4a+3\equiv 0\end{array}\bigg\rbrace \Rightarrow \left[\begin{array}{cc} 7 & 5 \\ 4 & 3\end{array}\right] \left[\begin{array}{c}\rm a \\ 1\end{array}\right]\equiv \left[\begin{array}{c}\rm 0 \\ 0\end{array}\right]\Rightarrow \left[\begin{array}{c}\rm a \\ 1\end{array}\right]\equiv \left[\begin{array}{c}\rm 0 \\ 0\end{array}\right]\Rightarrow 1\equiv 0\,\Rightarrow\, d\mid 1\!-\!0 $$
