recall your d = rt, distance = rate * time.
b = rate of the boat in still water
c = rate of the currrent
the distance going upstream is 8 miles, the distance going downstream is also the same 8 miles.
the boat took 1 hour going upstream, now, the boat is not going "b" mph fast, since it's going against the current, the current is eroding speed from, thus the boat going up is really going "b - c" fast.
likewise, when the boat goes downstream, is not going "b" fast either, is going faster because is going with the current and thus is really going "b + c" fast, and we know that trip back took 1/2 hour or 30 minutes.
[tex]\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Upstream&8&b-c&1\\ Downstream&8&b+c&\frac{1}{2} \end{array}\qquad \begin{cases} 8=(b-c)(1)\\ 8+c=\boxed{b}\\ \cline{1-1} 8=(b+c)\left( \frac{1}{2} \right) \end{cases}[/tex]
[tex]\bf \stackrel{\textit{substituting in the 2nd equation}~\hfill }{8=\left(\boxed{8+c}+c \right)\cfrac{1}{2}\implies 8=(8+2c)\cfrac{1}{2}}\implies 16=8+2c \\\\\\ 8=2c\implies \cfrac{8}{2}=c\implies \blacktriangleright 4=c \blacktriangleleft \\\\\\ \stackrel{\textit{we know that }~\hfill }{8+c=b\implies 8+4=b\implies \blacktriangleright 12=b \blacktriangleleft}[/tex]
