Is this right I need some help with this Please Alg1 ?

Question

Traveling against the wind, a plane flies 2100 miles from Chicago to San Diego in 4 hours and 40 minutes. The return trip, traveling with a wind that is twice as fast, takes 4 hours. Find the rate of the plane in still air.

Ok I got this

x+y=4 hours and 40 mintues (line 1)

2x-2y=4 hours(line 2) B/c it took twice as fast

Where does the 2100 miles come in at?

Answer 1

Any time you see a problem like this, you have to remember the old "rate times time equals distance" thing.  You're right that you're going to do some adding one way and some subtracting the other way, but it isn't time you're going to add.

Always define your variables before you start.  Believe me, it saves trouble later.  Let v be the speed of the plane in still air and w be the speed of the wind on the first leg.  Then we know 2w is the speed of the wind coming back.  Rate times time equals distance, so going we have

(v - w)(4 hours 40 minutes) = distance

It's v - w because you're flying against the wind, so that will tend to slow you down.

Now:  that equation is almost useless.  Let's make it more useful.  We need the time to be in a useful format, and "4 hours and 40 minutes" is not.  If we express it as hours and the distance as miles, we'll get miles per hour for the speed.  If we express it as minutes and the distance as miles, we'll get miles per minute.  Let's make it miles and hours.  So, the distance from San Diego to Chicago is 2100 miles.  4 hours 40 minutes is 4 hours + 40/60 hours = 4 2/3 hours = 14/3 hours

The equation, then, is

(v - w)(14/3) = 2100

On the return leg, the time is 4 hours.  That's easy.  We also know that the distance is the same, and that the speed of the wind is twice what it was, so

(v + 2w)(4) = 2100

That gives us two equations in two unknowns, so we can solve that.

Does that help?