Elimination Method in Real LifeDate: 03/30/2003 at 11:09:58 From: Cornelia Subject: Comparing Quantities If 3 tall candles and 5 small candles cost $7.30, and 2 tall candles and 2 small candles cost $3.40, how much does 1 tall candle cost? How much does 1 small candle cost? Date: 03/30/2003 at 12:14:35 From: Doctor Greenie Subject: Re: Comparing Quantities Hi, Cornelia - Let's think about this problem first as a real-life situation and see how we can determine the prices of each tall and each small candle; then we can look at the algebraic solution and see how it exactly parallels our reasoning in the real-life situation. We know that if a person goes into the store and buys 3 tall candles and 5 small candles, it will cost $7.30; and we know that if another person goes into the store and buys 2 tall candles and 2 small candles, it will cost $3.40. How can we use this information to find the cost of each small and each tall candle? Let's send several people into the store, having each of them buy either 3 tall and 5 small candles for $7.30 or 2 tall and 2 small candles for $3.40. And let's do it in such a way that, among the several people, the total number of tall candles (or small candles) is the same. When we do that, the total cost of the tall (or small) candles is the same, so the difference in the total cost represents the difference in the total number of small (or tall) candles that were bought. This allows us to determine the cost of the small (or tall) candle; and once we know that price we can determine the cost of the other. This explanation is perhaps a bit confusing. Let's clear it up by applying it to your particular example. For the information we have, in one group of candles there are 3 tall candles and in the other group there are 2. So let's send two people into the store, each of whom will buy 3 tall candles and 5 small candles for $7.30. Between them, they buy a total of 6 tall candles and 10 small candles for $14.60. And let's send three people into the store, each of whom will buy 2 tall candles and 2 small candles for $3.40. Between them, they buy a total of 6 tall candles and 6 small candles for $10.20. Each of the two groups of people bought a total of 6 tall candles. The difference between the total purchases of the two groups is then the difference in the number of small candles; the first group bought a total of 10 and the second group bought a total of 6, so the difference between the total purchases of the two groups is 4 small candles, and the difference in the total cost for the two groups was $14.60 - $10.20 = $4.40. So the cost of 4 small candles must be $4.40, and so the cost of each small candle is $1.10. Now that we know the cost of one small candle, from either of the original pieces of information we can determine the cost of a tall candle. For example, we know that 2 tall candles and 2 small candles cost $3.40; and we know the 2 small candles cost $2.20, so the two tall candles must have cost $1.20, which means each one cost $.60. Now let's look at the algebraic solution and see how similar it is to the solution described above. We start by defining the unknown quantities we are looking for: let t = cost of each tall candle let s = cost of each small candle Then we use these variables to write equations reflecting the known cost of the different combinations of tall and small candles; note that we write our equations in "cents" and without the decimal points and "$" symbols, because they just confuse us: 3t + 5s = 730 2t + 2s = 340 Now we look at the numbers of tall and small candles in the two combinations and we look for ways to multiply one or both of the equations by certain numbers so that the total numbers of tall (or small) candles is the same. With the "3s" in the first equation and the "2s" in the second, we multiply the first equation by 2 and the second by 3: 6t + 10s = 1460 6t + 6x = 1020 Now we subtract these two equations to find the difference between them: 6t + 10s = 1460 - (6t + 6s = 1020) ------------------- 4s = 440 Then we divide both sides of the equation by 4 to find 4s = 440 s = 110 And finally, we use this value which we have determine in one of the original equations to find the value of the other variable: 2t + 2s = 340 2t + 220 = 340 2t = 120 t = 60 This method of solving pairs of equations is called the eliminationn method, because we combine the given equations in such a way that one of the variables is eliminated. I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/