For this problem, we need the distance formula: rt=d.
The current is 2 km/h. Even though we don't know the rate of the boat in still water, we know that going upstream, the rate can be expressed as: (r-2). Downstream, the rate is (r+2).
Now let's deal with "time":
(t-upstream) + (t-downstream) = 10
Rewrite: (t-downstream) = 10 - (t-upstream)
now, we can use that second equation to replace (t-downstream) in a minute.
Getting back to the original formula: rt=d, we get 2 equations:
There is our system of equations. Let's solve it with substitution. We can solve the first equation in terms of "t":
Substitute that for "t" in the second equation. (This gets long-haired.)
Let's finally get rid of that complex fractions. Multiply every term by (r-2).
That bad boy is a quadratic equation, and I can think of no better tool than the quadratic formula. Hopefully you are already familiar with the quadratic formula, because instead of showing it, I just plugged it into the handy dandy solver (results below). It gave me 2 values, but one of them was negative. How can the boat go a negative rate? That makes no sense in this problem. We can throw that out. That leaves the positive answer, 3.09, as the rate.
If you plug that in, you'll see it works. Also, it took the boat 8.23 hours upstream, and 1.77 hours downstream. (Holy cow! It would have been MUCH faster to walk!)
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| Solved by pluggable solver: SOLVE quadratic equation with variable |
| Quadratic equation For these solutions to exist, the discriminant First, we need to compute the discriminant Discriminant d=1924 is greater than zero. That means that there are two solutions: Quadratic expression Again, the answer is: 3.09317121994613, -1.29317121994613. Here's your graph: |