Algebra 3 word problems!! Need desperate help!! PART 2!!?

Question

7. Megan and Heather rent a canoe for 2.5 hours. If they paddle out of the lake at 2 mph and return at 3 mph, what is the total distance they have canoed?

FOR PROBLEMS 8 AND 9: Robert has a jet ski that will travel at the rate of 35 mph in still water. He can go 50 miles up a river in the same amount of time it takes him to go 75 miles down the river.

8.How fast can Robert travel going down the river?
9. What is the total time required for robert to make his trip?

10. Allison and Rachelare taking a trip togther. They decide to share the driving respomsibilities for the 480 mile drive. Allison drove at an average rate of 50 mph and Rachel drove at an average rate of 40 mph. If Allison drove 45 minutes more than Rachel, then how far did Rachel drive?

Answer 1

For 7, you have two different variables involved - the time it took to go out and the time it took to come back in. You'll need to set up two equations for this one:

Distance covered:
2o = 3i (where o = time out, and i = time in - the distances are the same going out and coming back in)

Total time:
o + i = 2.5

You have two equations and two variables, so solve using whatever method you like. Substitution works for me:
2o = 3i
o = 3i/2

o + i = 2.5
3i/2 + i = 2.5
5i/2 = 2.5
5i = 5
i = 1

o = 3i/2
o = 3(1)/2
o = 3/2

So the distance is then 2o + 3i = 2(3/2) + 3(1) = 3 + 3 = 6 miles.

Questions 8 and 9 are also two equation situations, both based on distance = rate * time and accounting for the river current (r):

50 = (35 - r)t
75 = (35 + r)t

Again, substitute and solve:
50 = (35 - r)t
t = 50/(35 - r)

75 = (35 + r)t
75 = (35 + r)(50/(35 - r))
75(35 - r) = 50(35 + r)
2625 - 75r = 1750 + 50r
875 = 125r
r = 7 mph

t = 50/(35 - r)
t = 50/(35 - 7)
t = 50/28
t ~ 1.78 hours

So going down the river is the sum of his jet ski and the river current, or 35 + 7 = 42 mph, and the total time will be twice the time going one way, or 3.56 hours.

10 is again the same kind of problem, using distance = rate * time.

480 = 50a + 40r (a = Allison's time behind the wheel, and r = Rachel's time behind the wheel)

a = r + 0.75 (since Allison drove more, we add the 0.75 to Rachel, and we use 0.75 for 3/4 of an hour, or 45 minutes)

480 = 50a + 40r
480 = 50(r + 0.75) + 40r
480 = 50r + 37.5 + 40r
442.5 = 90r
r = 4.92 hours

a = r + 0.75
a = 4.92 + 0.75
a = 5.67 hours

Rachel thus drove 40r = 40(4.92) = 196.8 miles.

Answer 2

7.) Remember that v = d/t, right? They have to go out the same distance they come back, so we'll use d for the distance out, and 2d for total distance.

v = d/t
d = v1t1 (out) = v2t2 (in)
So: d = 2 mph * t1 = 3 mph * t2

Define t1 in terms of t2:
t1 + t2 = 2.5h
t1 = 2.5h - t2

Substitute:
d = 2 * (2.5 - t2) = 3 * t2
d = 5 - 2t2 = 3t2
5 = 5t2
t2 = 1 hr
t1 = 2.5h - t2
t1 = 1.5 hr

Now, plug in t1 and t2's value:
d = 2 mph * 1.5 = 3 mi

Total distance = 2d = 3mi * 2 = 6 mi (solution)

8. This is similar to #7:
v = d/t
t = d/v
t = d1/v1 = v2d2
t = 50/v1 = 75/v2

Define v1 and v2:
v1 = 35 mph - current (c)
v2 = 35 mph + c

Plug in:
t = 50/(35 - c) = 75/(35 + c)
t = 50/(35 - c) * (35 + c)/(35 + c) = 75/(35 + c) * (35 - c)/(35 - c)
t = 50(35 + c)/(35^2 - c^2) = 75(35 - c)/(35^2 - c^2)
t = 50(35 + c) = 75(35 - c)
t = 1750 + 50c = 2625 - 75c
125c = 875
c = 7 mph

Thus, Robert travels 35 mph + 7 mph down the river: 42 mph (solution)

9. For this, plug the value of c into t = 50/(35 - c) = 75/(35 + c). The total time is 2t.

10. This is done using the same method as 7 and 8.

Answer 3

This is the fourth desperate algebra question you have asked tonight. Try doing your own homework, instead of spending so much time trying to get someone else to do it for you.