Problem-Solving: Systems of Equations

Date: 4/27/96 at 20:16:45
From: Thomas Glasper
Subject: Systems of Equations

We have just started a new unit called "Systems of Equations" and I 
feel as if the whole class is going forward but me. I'm getting 
farther and farther behind. Could you help me with this problem? I 
don't even know where to begin.

Susan wants to have business cards printed.  One style will cost $25 
plus 2 cents per card.  Another style will cost $10 plus 5 cents per 
card.  For how many cards will the costs be the same for both?

Amber Green

Date: 5/4/96 at 16:35:25
From: Doctor Mike
Subject: Re:Systems of Equations

Hello Amber,   

In case it is not too late for you to benefit from an answer, here is 
the way I would approach this one:
  
I think it will be easiest to convert all money amounts to cents.  
Then, the first style costs "2500 + 2*N" cents for N cards.  
The second style costs "1000 + 5*N" cents for N cards.  
You want to find what N is when : 
   
          2500 + 2*N  =  1000 + 5*N
  
In most word problems like this, getting the equation is the hardest 
part. This is the clue you need to find out what N must be.  Remember 
that this equation will stay true if you do EXACTLY the same thing to 
both sides. 

So, here goes.  There are three steps: 
  
First, subtract 1000 from each side of the equation to get : 
   
          1500 + 2*N  =  5*N 
  
Second, subtract 2*N form each side of that equation to get : 
    
          1500  =  3*N
  
Third, divide each side of the equation by 3 to get : 
  
          500  =  N 
  
And that's your answer.  If the size of the order is 500 cards then 
the cost will be the same for either style.  You should go back and 
verify this by showing that for this size of order the cost is $35 
(3500 cents).

-Doctor Mike,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/