Rational expressions...?

Question

Could anyone do these and explain how they're done, please?

r/(3r + 6) = r/(5r + l0)

Use d = rt to solve: A long distance cyclist pedaling at a steady rate travels 30
miles with the wind. He can travel only 18 miles against the wind in the same amount of time.
If the rate of the wind is 3 miles per hour, what is the cyclist's rate without the wind?

Answer 1

On the first one, cross multiply to get 3r^2 + 6r = 5r^2 + 10r

This is 0 = 2r^2 + 4r = 2r(r+2)

So r could be 0 or -2, except that -2 causes the denominators to be zero.

So r=0 is the only answer

For the second, you have 2 equations from distance = rate x time:

30 = (r+3)t
18 = (r-3)t
where t = time and r = rate with no wind

30 = rt + 3t
18 = rt - 3t

subtract to get

12 = 6t so t = 2

Plug in to get r

Answer 2

Since the numerators on either side of the equation are the same, we may conclude that the denominators on both side of the equation have to be equivalent.

3r + 6 = 5r + 10
2r = -4
r = -2

but r = -2 cannot be a solution since it will make both denominators zero.

Then the only option is when the numerators are both zero to make this equation true.

As for the second part of your question, I think you should not put unrelated problems under one question.

Let me give you a hint: rate with the wind (r + 3) --- faster
rate against the wind (r - 3) ---- slower

Plug in d = rt and solve for r.

Answer 3

What do these two problems have to do with each other?