Respuesta :
recall your d = rt, distance = rate * time.
going forth and back is 80 miles, thus upstream as well as downstream is just 80 miles.
now, bear in mind that say, the boat has a "still water" of "b", when going downstream, is not going "b" fast is going "b+7" because the current's is adding to it, likewise when is going upstream is going "b-7" because is going against the current and thus the current is eroding speed from it.
now, the whole trip took 3 hours and 20 minutes, bearing in mind that 20 minutes is 1/3 of an hour, so is 3 and 1/3 hours, or 10/3 of hours.
if say it took "t" hours going down, then going up it took the slack from 10/3 and "t", that is, it took "10/3 - t".
[tex]\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ Downstream&80&b+7&t\\ Upstream&80&b-7&\frac{10}{3}-t \end{array} \\\\\\ \begin{cases} 80=t(b+7)\implies \frac{80}{b+7}=\boxed{t}\\\\ 80=(b-7)\left(\frac{10}{3}-t \right)\\ ----------\\ 80=(b-7)\left(\cfrac{10}{3}- \boxed{\frac{80}{b+7}} \right) \end{cases}[/tex]
[tex]\bf \cfrac{80}{b-7}=\cfrac{10}{3}-\cfrac{80}{b+7}\implies \cfrac{80}{b-7}=\cfrac{10(b+7)-80(3)}{3(b+7)} \\\\\\ \cfrac{80}{b-7}=\cfrac{10b+70-240}{3(b+7)}\implies \cfrac{80}{b-7}=\cfrac{10b-170}{3(b+7)} \\\\\\ 3(b+7)80=(b-7)(10b-170)\\\\\\ 240(b+7)=(b-7)(10b-170) \\\\\\ 240b+1680=10b^2-240b+1190\implies 0=10b^2-480b-490 \\\\\\ 0=b^2-480b-49\implies 0=(b-49)(b+1)\implies b= \begin{cases} \boxed{49}\\ -1 \end{cases}[/tex]
going forth and back is 80 miles, thus upstream as well as downstream is just 80 miles.
now, bear in mind that say, the boat has a "still water" of "b", when going downstream, is not going "b" fast is going "b+7" because the current's is adding to it, likewise when is going upstream is going "b-7" because is going against the current and thus the current is eroding speed from it.
now, the whole trip took 3 hours and 20 minutes, bearing in mind that 20 minutes is 1/3 of an hour, so is 3 and 1/3 hours, or 10/3 of hours.
if say it took "t" hours going down, then going up it took the slack from 10/3 and "t", that is, it took "10/3 - t".
[tex]\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ Downstream&80&b+7&t\\ Upstream&80&b-7&\frac{10}{3}-t \end{array} \\\\\\ \begin{cases} 80=t(b+7)\implies \frac{80}{b+7}=\boxed{t}\\\\ 80=(b-7)\left(\frac{10}{3}-t \right)\\ ----------\\ 80=(b-7)\left(\cfrac{10}{3}- \boxed{\frac{80}{b+7}} \right) \end{cases}[/tex]
[tex]\bf \cfrac{80}{b-7}=\cfrac{10}{3}-\cfrac{80}{b+7}\implies \cfrac{80}{b-7}=\cfrac{10(b+7)-80(3)}{3(b+7)} \\\\\\ \cfrac{80}{b-7}=\cfrac{10b+70-240}{3(b+7)}\implies \cfrac{80}{b-7}=\cfrac{10b-170}{3(b+7)} \\\\\\ 3(b+7)80=(b-7)(10b-170)\\\\\\ 240(b+7)=(b-7)(10b-170) \\\\\\ 240b+1680=10b^2-240b+1190\implies 0=10b^2-480b-490 \\\\\\ 0=b^2-480b-49\implies 0=(b-49)(b+1)\implies b= \begin{cases} \boxed{49}\\ -1 \end{cases}[/tex]
