Word problem for my Algebra 1 class!!

Question

Flying against the wind, an airplane travels 5340 km in 6 hours. Flying with the wind, the same plane travels 11,250 km in 9 hours. What is the rate of the plane in still air AND what is the rate of the wind?

Answer 1

So this problem requires that we solve a system of equations. First, we must identify what the end rate is in each case, this is just the distance traveled in each case divided by the number of hours:

r_a = 5340km/6hr = 890 km/hr (for against the wind)

r_p = 11250km/9hr = 1250 km/hr (for with the wind)

Now with this information the rate of travel should be the still travel rate plus or minus the wind rate depending on whether we are traveling with the wind or against the wind respectively. so we have:

r_s + r_w = r_p

and

r_s - r_w = r_a

r_s is the still air rate and r_w is the wind rate

from this we can solve for r_s or r_w either by adding the equations or solving for one variable and plugging in. I'm going to add the equations together (so in columns just add, r_s + r_s + r_w -r_w = r_p + r_a):

2r_s = r_p + r_a = 2140 km/hr

We can solve this for r_s and then plug that solution into either of the above equations to solve for r_w.

Let me know if this is unclear or if you would like to see the other method.