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(male narrator) In this video,
we will look at setting up distance problems.
In Part 2 of this video,
we'll go back and solve the distance problems we set up.
To help us organize the information we're given
into a simultaneous product, we're gonna set up our table
knowing that distance is equal to rate,
times the time, gives us the distance.
This'll give us a simultaneous product,
which we can solve by dividing by what...you want.
Let's take a look at some examples that we can set up,
and then in another video, we will solve.
In this video,
a man's riding his bike to the park 60 miles away.
We know rate times time, equals distance,
but we don't know the rate he rode in his bike,
and we don't know the time it took him to get there.
All we know is the distance was 60 miles away.
We are told his return trip, he went 2 miles per hour slower.
This means the rate has decreased by 2,
and as a result, the time took 1 hour longer.
The time is increased by 1.
It is still the same distance, and so, we see we end up
with a simultaneous product: rate times time, equals 60;
and rate minus 2, times time, plus 1, equals 60.
In order to start solving,
we will divide by what we're looking for.
The question to ask:
How fast did he ride to the park?
So we will divide by the rate on both equations.
This gives us: time is equal to 60 over r;
and dividing the other equation by r minus 2...
gives us time, plus 1, equals 60 over r, minus 2.
We are now ready to make a substitution
with the time into our equation.
Replacing the time with 60 over r, plus 1,
will be equal to 60 over r, minus 2.
And this gives us a rational equation we can solve
to find out how fast he rode to the park.
We will come back and solve this equation
in Part 2 of this video.
For now, let's take a look at another example.
Here, a woman is driving through a construction zone
for 45 miles.
We know rate times time, equals distance.
And we do not know the rate she drove,
or the time she drove that rate,
but we do know the distance she covered was 45 miles.
She then realized
that if she had driven 6 miles per hour faster,
this means she is increasing her rate by 6: r plus 6.
She would have arrived 2 hours sooner.
This is our time, minus 2, equals the distance.
She is still considering
the same distance she actually traveled,
and so, that distance is still 45.
You can see we have a simultaneous product
out of our table that rate times time, equals 45;
and rate plus 6, times time, minus 2, equals 45 as well.
In order to solve these equations,
we divide by the factor we're looking for.
The question's interested in how fast she drove,
so we will divide by the rate.
Dividing the first equation by the rate
tells us that time is equal to 45 over r.
Dividing the second equation by the rate factor--
or r plus 6--tells us that time minus 2
is equal to 45 over r, plus 6.
We can now bring the two equations together
by making a substitution
and replacing time in the second equation with the 45 over r.
When we do this,
we get 45 over r, minus 2, equals 45 over r, plus 6.
We now have a rational equation
that we could go through and solve.
We'll go back and solve both of these problems
in Part 2 of this video.