ALGEBRA QUESTION??!!!?

You drove 180 miles at a constant rate of speed.

If you had driven 15 miles per hour faster,

you would have traveled the same distance in an hour less.

How fast did you drive?

180/x - 180/(x + 15) = 1

180(x + 15) - 180x = x^2 + 15x

x^2 + 15x - 2700 = 0

(x - 45)(x + 60) = 0

x = 45, -60

I drove at 45 miles per hour.

Set up your two equations:

R1:

Speed = x

Distance = 180

Time = y

R2:

Speed = x + 15

Distance = 180

Time = y - 1

Now form your two system of equations (using speed = distance/time):

x = 180/y

x + 15 = 180/(y - 1)

Substitute 180/y for x in the second equation:

180/y + 15 = 180/(y - 1)

Multiply everything by y(y - 1) to cancel off all denominators:

180(y - 1) + 15y(y - 1) = 180y

Expand all brackets:

180y - 180 + 15y^2 - 15y = 180y

Move everything to one side:

15y^2 - 15y - 180 = 0

Divide everything by 15 to simplify the equation:

y^2 - y - 12 = 0

Factor it out:

(y - 4) (y + 3) = 0

This gives you the possibility of the time you travelled in being either 4 hours or -3 hours, obviously there is no such thing as a negative time, so the time you took can only be 4:

Speed = x

Distance = 180

Time = 4

Speed = Distance/Time

x = 45

You drove at a speed of 45 km/ph.

"You drove 180 miles at a constant rate of speed (s)"

time = distance/rate

t = 180/s

"If you had driven 15 miles per hour faster, you would have traveled the same distance in an hour less."

t - 1 = 180/(s + 15)

(180/s) - 1 = 180/(s + 15)

(180 - s)/s = 180/(s + 15)

(180 - s)(s + 15) = 180s

(s - 180)(s + 15) = -180s

s² - 165s - 2700 = -180s

s² + 15s - 2700 = 0

quadratic formula

s = [-15 ± √(15² – 4·1(-2700))] / [2·1]

   = [-15 ± √11025] / 2

   = [-15 ± 105] /2

   = -60, 45 s can't be negative, so s = 45 miles per hour.