Expert: Bobby Soltani - 2/26/2005

Question
Hey there Mr.Soltani!Pls. try to anwer my problem ASAP. Ok, this is the actual problem:

Music A Club has a membership fee of $30 and charges $8 per CD. Music B Club has a membership of $45 and charges $7 per CD. How many CD's would a person need to buy for club A to be more expensive than CLub B?

I hope you explain it to me very well. Here is some info:
Book:Algebra 1 by Mcdougal Littell
My grade:9
Topic/Lesson:About solving a system of linear equations by different methods(graphing,substituting, and linear combinations).

Lots of Thanks,
Clyde

Answer
Hi Clyde,

There are several ways to solve this problem.  Here is one.

The cost for A is
CA = 30 + 8n , where CA is the cost for club A and n is the number of CDs purchased.

The cost for B is
CB = 45 + 7n

we can see that when the number of cds is 0, A cost 30 and b costs 45.  But, since the CDs are cheaper at B, if you buy enough CDs, it will become cheaper.  If we find out when the costs are equal to eachother, we will know when B becomes cheaper.  To do that, let's set the two equations equal to each other and solve for n

30 + 8n = 45 + 7n
subtract 30 from both sides
8n = 15 + 7n
subtract 7n from both sides
n = 15

That means that the two clubs will cost the same if you buy 15 cds.  If you buy more than 15, then A becomes more expensive than B.

Let me know if you have any questions.

Bobby