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/
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