SYSTEM OF EQUATION PROBLEM HELPPPP

Question

A jet travels 1887 mi against the wind in 3 hours and 2157 mi with the wind in the same amount of time what is the rate of the jet in still air and what is the rate of the wind

Answer 1

The distance formula is d = rt     so    r = d/t

d = distance, t = time, r = rate

 

Let J = rate of Jet in still air

Let W = rate of Wind

J-W = rate of Jet against the wind

J+W = rate of Jet with the wind

 

The Jet flies 1887 miles against the wind in 3 hours so:  J-W = 1887/3 = 629 mph

The Jet flies 2157 miles with the wind in 3 hours so:      J+W = 2157/3 = 719 mph

 

The speed of the Jet is the average of the two rates

J = (629 + 719)/2 = 674 mph

 

J-W = 629 so  674-W = 629

                     W = 674-629 = 45 mph

 

The Jet in still air flies at a rate of 674 mph

The wind speed is 45 mph

Answer 2

Let j = rate of jet

Let w = rate of wind

 

1887 = 3(j-w)

2157 = 3(j+w)

 

Subtract

270 = 6w

w = 45 miles/hr

3j=1887+135

j=2022 miles/hr

 

Answer 3

You need to use this formula

 

Distance = Rate x Time.

 

Let d be distance, r be rate, and t be time.

 

So you know that the jet travels 1887 miles against the wind in 3 hours.  Plugging this into the formula we get

 

1887 = rate against the wind x 3

 

We also have that the jet travels 2157 miles with the wind in 3 hours.  Plugging this into the formula we get

 

2157 = rate with the wind x 3

 

Let r_j be the rate of the jet in still air (alone) and let r_w be the rate of the wind. The rate against the wind means that the wind is making the jet move slower and the rate with the wind means that the wind is making the jet move faster.

 

In other words, we have that the rate against the wind is equal to r_j minus r_w and the rat with wind is equal to  r_j plus r_w.

 

Then we have

 

1887 = (r_j - r_w) x 3

2157 = (r_j + r_w) x 3

 

You now have a system of equations and can solve for r_j and r_w.