Question 181288This question is from textbook Fourth Edition Elementary and Intermediate Algebra
: 1. Steve traveled 150 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 2 hours less. Find the speed of his vehicle.
This question is from textbook Fourth Edition Elementary and Intermediate Algebra
Answer by solver91311(24713)   (Show Source):
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Consider the basic formula that relates distance, rate (speed), and time:
and its variations:
and
It will be convenient to use the last one because that will ultimately allow us to solve for the desired quantity, r, directly.
The actual trip can be described thus:
And the 'what if' (20 mph faster, 2 hours less) trip can be described:
 + 2 \ \ \Rightarrow \ \ t = \left(\frac{150}{r + 20}\right) + \left(\frac {2(r + 20)}{r + 20}\right) \ \ \Rightarrow \ \ t = \frac {2r + 190}{r + 20} )
Now we have expressed the variable t in two different ways in terms of r, the quantity we seek. So set these two expressions in r equal to one another.
This is a simple proportion that can be solved by first cross-multiplying and simplifying:
 = 2r^2 + 190r \ \ \Rightarrow \ \ -2r^2 -40r + 3000 = 0 \ \ \Rightarrow \ \ r^2 + 20r - 1500 = 0)
Leaving us with a factorable quadratic:
Since  and  , we can say:
(r + 50) = 0)
Hence,
or
But -50 for a speed is absurd and is therefore an extraneous root introduced by the action of squaring the variable during the solution process. Exclude -50.
That leaves
as the answer.
Check:
If r were 20 mph faster, then  and if t were 2 hours less, then  . If the solution is correct, then a trip lasting 3 hours at 50 miles per hour should cover the same 150 mile distance.
Answer checks.
John
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