Question

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?

Answer 1

Let x be the time it takes for the trip to be completed given that the speed is y.

When the time is 30 minutes (equal to 0.5 hour) shorter than x, the speed is 8 mph more than the original speed. 

The equations that would best represent the given conditions are:
      (1)             120 = (x)(y)
      (2)              120 = (x - 0.5)(y + 8)

Simplifying,
                       y = 120/x
Substitute:
                    120 = (x - 0.5)(120/x + 8)

The value of x from the equation is 3. Thus, if I go with the usual speed, the time it will take me to finish the trip is approximately 3 hours. 

Answer 2

Answer: 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:

$t = \frac{4 +- \sqrt{16 +4*60*8} }{16} = \frac{4+-\sqrt{1936} }{16}  = \frac{4 +-44}{16}$

So we have two solutions: $t = \frac{4+44}{16} = 3h$ and $t = \frac{4 -44}{16} = -2.5h$.

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.