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Translate to a system of equations and then solve: A Mississippi river boat cruise sailed 120 miles upstream for 12 hours and then took 10 hours to return to the dock. Find the speed of the river boat in still water and the speed of the river current.

The rate of the boat is 11 mph and the rate of the current is 1 mph.

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Translate to a system of equations and then solve: Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.

The speed of the canoe is 7 mph and the speed of the current is 1 mph.

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Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in [link] . A wind current in the same direction as the plane is flying is called a tailwind . A wind current blowing against the direction of the plane is called a headwind .

Translate to a system of equations and then solve:

A private jet can fly 1095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

Read the problem.

This is a uniform motion problem and a picture will help us visualize.
This figure shows an arrow labeled “3 hours” which continues to the right, representing the wind. Under the wave is a ray that points to the right and is labeled “j plus w equals 365” and “1,095 miles”. Under this ray is another ray pointing to the left labeled “j minus w equals 329” and “987 miles.”

Identify what we are looking for. We are looking for the speed of the jet
in still air and the speed of the wind.
Name what we are looking for. Let j = the speed of the jet in still air.
w = the speed of the wind
A chart will help us organize the information.
The jet makes two trips-one in a tailwind
and one in a headwind.
In a tailwind, the wind helps the jet and so
the rate is j + w .
In a headwind, the wind slows the jet and
so the rate is j w .
.
Each trip takes 3 hours.
In a tailwind the jet flies 1095 miles.
In a headwind the jet flies 987 miles.
Translate into a system of equations.
Since rate times time is distance, we get the
system of equations.
.
Solve the system of equations.
Distribute, then solve by elimination.
.
Add, and solve for j .

Substitute j = 347 into one of the original
equations, then solve for w .
.
.
.
.
.
Check the answer in the problem.

 With the tailwind, the actual rate of the
 jet would be
   347 + 18 = 365 mph.
 In 3 hours the jet would travel
    365 · 3 = 1095 miles.
 Going into the headwind, the jet’s actual
 rate would be
   347 − 18 = 329 mph.
 In 3 hours the jet would travel
    329 · 3 = 987 miles.
Answer the question. The rate of the jet is 347 mph and the
rate of the wind is 18 mph.

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Translate to a system of equations and then solve: A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1035 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

The speed of the jet is 235 mph and the speed of the wind is 30 mph.

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Translate to a system of equations and then solve: A commercial jet can fly 1728 miles in 4 hours with a tailwind but only 1536 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

The speed of the jet is 408 mph and the speed of the wind is 24 mph.

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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