Math help?

Question

A 1,000 mile trip against the wind took 4 hours. The same trip with the wind took 3 hours and 20 minutes. How fast is the wind, and how fast was I going?

(solve using a system of two equations)

Answer 1

You'd be using the equation d = vt (distance = velocity * time)

Let x = your speed (miles per hour), and w = windspeed (mph)

Travelling against the wind, your velocity would be x-w; travelling with the wind your velocity would be x+w, so:

(x-w)*4 hours = 1000 miles
(x+w)*3-1/3 hours = 1000 miles

4x - 4w = 1000 and
3.333x + 3.333w = 1000

multiply the second equation by 4/3.333 to get:

4x - 4w = 1000
4x + 4w = 1200

add the two equations to get:

8x = 2200, and thus x = 275 mph

now solve for w using either equation:

(4*275) - 4w = 1000
-4w = -100
w = 25 mph

Answer 2

This is a system of linear equations. You have to create two equations with two unknowns and then solve for x and y.

Let x be the speed of the person walking.
Let y be the speed of the wind.

Converting the time into minutes, the two equations are:

1000 / 240 = x - y
1000 / 200 = x + y

Rearranging the first equation we get
x = 1000 / 240 + y

Substituting this into the second equation we get
1000 / 200 = (1000 / 240 + y) + y
1000 / 200 - 1000 / 240 = 2y
5/6 = 2y
5/12 = y

Now we subsitute y back into the first equation.

1000 / 240 = x - 5/12
1000 / 240 + 5 / 12 = x
4 7/12 = x

Therefore the wind was going at 5/12 miles per minute, or 25 miles/hour, and the person was travelling at 4 7/12 miles per minute or 275 miles/hour.

Answer 3

wind made difference of 40 min or 2/3 hr
1000/(s-w)=4
1000=4s-4w
250=s-w

1000/(s+w)=10/3
1000=10s/3+10w/3
300=s+w add 2 eqns
250=s-w
550=2s
s=275 mi/hr
s+w=300
275+w=300
w=25 mi/hr

Answer 4

you 300 .Wind 50

Answer 5

not enough information you either have to have the wind speed or something else srry