A car travels 133 mi averaging a certain speed. If the car had gone 30 mph faster, the trip would have taken 1 hr less. Find the car's average speed.

49.923 mph

Step-by-step explanation:

we know that the car traveled 133 miles in h hours at an average speed of x mph.

That is, xh = 133.

We can also write this in terms of hours driven: h = 133/x.

If x was 30 mph faster, then h would be one hour less.

That is, (x + 30)(h - 1) = 133, or h - 1 = 133/(x + 30).

We can rewrite the latter equation as h = 133/(x + 30) + 1

We can then make a system of equations using the formulas in terms of h to find x:

h = 133/x = 133/(x + 30) + 1

133/x = 133/(x + 30) + (x + 30)/(x + 30)

133/x = (133 + x + 30)/(x + 30)

133 = x*(133 + x + 30)/(x + 30)

133*(x + 30) =  x*(133 + x + 30)

133x + 3990 = 133x + x^2 + 30x

3990 = x^2 + 30x

x^2 + 30x - 3990 = 0

Using the quadratic formula:

x = [-b ± √(b^2 - 4ac)]/2a  

= [-30 ± √(30^2 - 4*1*(-3990))]/2(1)  

= [-30 ± √(900 + 15,960)]/2

= [-30 ± √(16,860)]/2

= [-30 ± 129.846]/2

= 99.846/2    x is miles per hour, and a negative value of x is neglected, so we'll use the positive value only)

= 49.923

Check if the answer is correct:

h = 133/49.923 = 2.664, so the car took 2.664 hours to drive 133 miles at an average speed of 49.923 mph.

If the car went 30 mph faster on average, then h = 133/(49.923 + 30) = 133/79.923 = 1.664, and 2.664 - 1 = 1.664.

Thus, we have confirmed that a car driving 133 miles at about 49.923 mph would have arrive precisely one hour earlier by going 30 mph faster

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