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rcar1029 · Sep 22, 2010, 03:38 PM

Unknown008 · Sep 23, 2010, 02:15 AM

Could you repost your question in the answer box please? Titles cannot support all the characters in your question.

rcar1029 · Sep 23, 2010, 05:04 AM

a salesperson drove the first 105 miles of a trip in 1 hour more than it took to drive the last 90 miles. The average rate during the last 90 miles was 10mph faster than the average rate during the first 105miles. Find the average rate for each portion of the trip(quadratic equations)

Unknown008 · Sep 23, 2010, 05:14 AM

Ok, set up equations. Let's call the first speed he used for the first part of the trip x mil/hr and the other part of the trip y mil.hr.

From the first sentence,

$Speed \times Time = Distance$

$x(t+1) = 105$

$y(t) = 90$

where t is the time he took to cover the last 90 miles.

From there, we can find x in terms of y by making t the subject of formula.

$x(t+1) = 105$

$t = \frac{105}{x} - 1$

$y$\frac{105}{x} - 1$ = 90$

$105y = 90x+ xy$ ... (I)

From the second part of the question, we know that y is 10 more than x.

So,

$y =x+10$ ... (II)

You have two equations with two variables. You can solve for x and y now, the average speed the salesman took for the 105 miles part and 90 miles part.

Post what you get! :)