Your answer is correct. You can use algebra to show that 11 4-hops and 2 3-hops is the only possibility. In fact, suppose $x$ is the number of 4-hops...
\$...\$ gives \textstyle, but ... produces \displaystyle. You can see the difference if you set the formula \lim_{x\to0}\frac{f(x)}{g(x)} in the two...
Almost. You can think of the minus sign as canceling: $|a|=a$. But now, whether there's a solution or not depends greatly on $a$. If $a>0$, there is...