SOLUTION: Two cars leave an intersection. One car travels north; the other travels east. When the car traveling north had gone 24 miles, the distance between the cars was four miles more t

Question 248502: Two cars leave an intersection. One car travels north; the other travels east. When the car traveling north had gone 24 miles, the distance between the cars was four miles more than three times the distance traveled by the car heading east. What was the distance between the cars at that time?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Two cars leave an intersection. One car travels north; the other travels east.
When the car traveling north had gone 24 miles, the distance between the cars was four miles more than three times the distance traveled by the car heading east.
What was the distance between the cars at that time?
:
This is a pythag problem: a^2 + b^2 = c^2
a = 24 mi (north bound car dist)
b = eastbound car dist
c = dist between the cars
:
It says," the distance between the cars was four miles more than three times the distance traveled by the car heading east." therefore
c = 3b + 4
:
24^2 + b^2 = (3b+4)^2
:
FOIL the right side
576 + b^2 = 9b^2 + 24b + 16
:
A quadratic equation, find b
0 = 9b^2 - b^2 + 24b + 16 - 576
:
8b^2 + 24b - 560 = 0
;
Simplify, divide by 8
b^2 + 3b - 70 = 0
Factor
(b+10)(b-7) = 0
:
b = 7 mi
:
c = 3(7) + 4
c = 25 mi is the dist between the cars
;
:
Is this true?
24^2 + 7^2 = 25^2
576 + 49 = 625; seems to be.

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