My trip to work is 120 miles. if i go 8 mph faster than my usual speed, i'll get to work 30 minutes earlier. how long does my trip take, in hours, if i go my usual speed?

Hello!

The total distance is 120 miles, and you know that if you go 8 mi/h faster than usual you get there 30(or 0.5 hours) minutes early.

So if v is your usual speed, and t is your usual time, we have the next equations:

1) v*t = 120mi

2) (v + 8mi/h)*(t - 0.5h) = 120 mi

In equation (1) we can write v as a function of t; this is v = 120mi/t, and replace it in the second equation.

(v + 8)*(t - 0.5) = 120

(120/t + 8)(t - 0.5) = 120

120 + 8*t -60/t - 4 = 120

8*t -60/t - 4 = 0

now we need to obtain the value of t. Multiplying by t in both sides we have:

8*t^2 -60 - 4t = 0

Now we can use Bhaskara to obtain the two possible values for t:

So we have two solutions: and .

The second is a negative time, so this has no sense; then we only took the first solution; when you go at your usual speed, your trip takes 3 hours.

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