Question 989537: Pete and Repeat are brothers. Of course with names like Pete and Repeat they usually quarrel when it comes to making decisions. The brothers have decided they would like to open a small neighborhood pet store. After many heated discussions they have decided to sell dogs, cats, and exotic fish(no turtles or birds). Together they have a $1000 to spend on animals for their venture. Pete thinks they have room for 100 animals in the store. Repeat knows that cats cost $5, dogs cost $10 and exotic fish cost $100(they are really unusual and large fish). "Pete declares we must buy a total of 100 Animals" Repeat argues back "But we need to spend all our money to get the best selection for our customers" Help the brothers determine how to stock their new pet store and make both brothers happy with their decision.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
There are six different solutions if you allow the case where one or more of the animal types are not represented at all, 5 valid solutions otherwise.
Let  represent the number of cats,  represent the number of dogs,  represent the number of fish.
Solve this by cases, where takes on values in the range 0 through 10. If you wish to disallow the case where any animal type is not represented at all, then start at 1. For each value of z you choose, adjust the value of the constant term and then solve the 2X2 system for and . At some point, you will start getting a negative number for the variable, so you can discard that value of and anything larger.
John
My calculator said it, I believe it, that settles it
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