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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 -

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

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Equations
Middle School Word Problems

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