Answer:
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
<em><u>Using the quadratic formula: </u></em>
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
