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Elimination Method in Real Life

Date: 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/ 
Associated Topics:
Middle School Equations
Middle School Word Problems

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