Respuesta :
so... hmm bear in mind, when the boat goes upstream, it goes against the stream, so, if the boat has speed rate of say "b", and the stream has a rate of "r", then the speed going up is b - r, the boat's rate minus the streams, because the stream is subtracting speed as it goes up
going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r
notice, the distance is the same, upstream as well as downstream
thus [tex]\bf \begin{cases} b=\textit{rate of the boat}\\ r=\textit{rate of the river} \end{cases}\qquad thus \\\\\\ \begin{array}{lccclll} &distance&rate&time(hrs)\\ &----&----&----\\ upstream&48&b-r&4\\ downstream&48&b+4&3 \end{array} \\\\\\ \begin{cases} 48=(b-r)(4)\to 48=4b-4r\\\\ \frac{48-4b}{-4}=r\\ --------------\\ 48=(b+r)(3)\\ -----------------------------\\\\ thus\\\\ 48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3) \end{cases}[/tex]
solve for "r", to see what the stream's rate is
what about the boat's? well, just plug the value for "r" on either equation and solve for "b"
going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r
notice, the distance is the same, upstream as well as downstream
thus [tex]\bf \begin{cases} b=\textit{rate of the boat}\\ r=\textit{rate of the river} \end{cases}\qquad thus \\\\\\ \begin{array}{lccclll} &distance&rate&time(hrs)\\ &----&----&----\\ upstream&48&b-r&4\\ downstream&48&b+4&3 \end{array} \\\\\\ \begin{cases} 48=(b-r)(4)\to 48=4b-4r\\\\ \frac{48-4b}{-4}=r\\ --------------\\ 48=(b+r)(3)\\ -----------------------------\\\\ thus\\\\ 48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3) \end{cases}[/tex]
solve for "r", to see what the stream's rate is
what about the boat's? well, just plug the value for "r" on either equation and solve for "b"
