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girlongixxer
07-06-2006, 07:33 PM
Is there such thing as 3 variables in algebra?

pka
07-06-2006, 07:41 PM
If N is a counting number then there can be N variables.
So you need to answer this question: "Is 3 a counting number?"

girlongixxer
07-06-2006, 09:15 PM
I think what I meant to say was what if an equation had three variables? I have been working with equations that had me find out what x equals and what y equals but not an equation with another added variable. Can it be done? Is there any such thing?

stapel
07-06-2006, 09:25 PM
An equation can have any number of variables.

One would not be able to solve for the values of the variables, given one equation with more than one variable.

One can easily solve a system of three linear equations in three variables. A non-linear system is usually harder.

Eliz.

Denis
07-06-2006, 10:39 PM
go here, gixxer:

http://mathforum.org/library/drmath/view/57643.html

girlongixxer
07-07-2006, 08:17 AM
This is sooooooo confusing. But thank you anyway.

tkhunny
07-07-2006, 08:36 AM
I think what I meant to say was what if an equation had three variables? I have been working with equations that had me find out what x equals and what y equals but not an equation with another added variable. Can it be done? Is there any such thing?I dare you to describe the environment around you with only two variables. What is the size of your house?

girlongixxer
07-07-2006, 08:58 AM
My house is 1362 square feet.

Denis
07-07-2006, 10:35 AM
My house is 1362 square feet.

That's probably the area of the rectangle formed by the base of the house;
that's NOT the size of the house: what if it's only a 1 floor house, height 6 feet

And why do you find it "sooooo confusing"?
It's simply a little more difficult than a 2 variable.

In cents: apples = 15 each, bananas = 20 each, cantaloupes = 60 each;
you buy 9 apples, 12 bananas and 3 cantaloupes:
9a + 12b + 3c = 555 cents (or 5 dollars and 55 cents); see that?

You'll find out that you need 3 equations to be able to solve 3 variables:
similarly to needing 2 equations to solve 2 variables.

WHY are you having so much trouble with 3 variables: missed the
classes when this was being taught?

girlongixxer
07-07-2006, 10:50 AM
We are learning this in class now. That is why I came here to find clarity. Something you find simple - others find hard. That's all.

tkhunny
07-07-2006, 11:02 AM
My house is 1362 square feet.That is the horizontal surface area, possibly split amongst several surfaces (floors), as defined by various rules of realty. Good. You have defined two variables, Length and Width. That's how square feet is calculated.

How tall is it?

TchrWill
07-07-2006, 03:38 PM
[quote="girlongixxer"]Is there such thing as 3 variables in algebra?

As already stated, you can solve N equations with N variables.

SUrprisingly, you can solve one equation with two variables or one equation with three variables.

Example 1

In how many different ways can five-cent and eight-cent stamps be used to make \$2.43? What are the combinations?
This problem boils down to the solution of the equation .05x + .08y = 2.43 or, as I chose to use it, 5x + 8y = 243, making it easier to work with whole numbers. Now, as is typical with problems of this type, x and y must be positive whole numbers, a condition vital to the solution. Since 5 is the smallest of the two coefficients, we divide through by 5 and get x + y + 3y/5 = 48 + 3/5 which becomes (3y - 3)/5 = 48 - x - y. Now, (48 - x - y) must be a whole number, so (3y - 3)/5 must also be a whole number. We want to reduce this expression such that y has no coefficient so we multiply (3y - 3)/5 by 2 and get (6y - 6)/5, still a whole number. Dividing out by 5 yields y + y/5 - 1 - 1/5 = j, some unspecified whole number. Then, y/5 -1/5 = (y - 1)/5 = j - y + 1. Again, both sides are, by definition, whole numbers, so set (y - 1)/5 = k, another unspecified whole number. Then y = 5k + 1. Substituting this expression for y back into the original equation we have 5x + 40k + 8 = 243 or 5x = 235 - 40k or x = 47 - 8k. We now have two expressions for x and y in terms of k with which we can determine x and y for various values of k from zero to whatever. By inspection, it is clear that k must be less than six or x would be negative and we are looking for positive whole number answers. So the only values of k that can even be considered are from 0 to 5. In tabular form we get--
----k 0 1 2 3 4 5 6 7
....x 47 39 31 23 15 7 neg neg (no more valid answers)
....y 1 6 11 16 21 26 -- --
These are therefore the only six combinations that will result in integer answers. Going through the arithmetic, you will see that all the combinations all add up to \$2.43.

Example 2

Joe came back from a stamp collectors gathering and told his sister, Jill, that he bought a hundred stamps. Joe said he bought the stamps at four different prices: \$0.59, \$1.99, \$2.87, and \$3.44 each. Jill asked, "How much did you pay altogether?" Joe replied, "One hundred dollars exactly." How many stamps did Joe buy at each
price?

Given: (1)--W + X + Y + Z = 100 and (2)--.59W + 1.99X + 2.87Y + 3.44Z = 100.
1--Multiply (2) by 100--->59W + 199X + 287Y + 344Z = 10000. (3)
2--Multiply (1) by 59----->59W + 59X + 59Y + 59Z = 5900. (4)
3--Subtract (4) from (3)--->140X + 228Y + 285Z = 4100. (5)
4--Divide (5) through by 140--->X + Y + 88Y/140 + 2Z + 5Z/140 = 29 + 40/140.
5--Solving for X = 29 - Y - 2Z - (88Y + 5Z - 40)/140. (6)
6--Set (88Y + 5Z - 40)/140 = u = an integer.
7--Rearranging, 140u = 88Y + 5Z - 40. (7)
8--Dividing (7) through by 5--->28u = 17Y + 3Y/5 + Z - 8. (8)
9--Solving for Z, Z = 28u - 17Y - 3Y/5 + 8. (9)
10--With 3Y/5 = integer, multiply by 2 yielding 6Y/5. (10)
11--Dividing (10) out yields 6Y/5 = Y + Y/5 where Y/5 must be an integer also.
12--Set Y/5 = v whence Y = 5v. (11)
13--Substituting (11) into (8)--->Z = 28u - 85v - 3v + 8 = 28u - 88v + 8. (12)
14--Substituting (11) and (12) into (6)--->X = 29 - 5v -56u + 176v -16 - (440v + 140u - 440v + 40 - 40)/140. (13)
15--Simplifying (13)--->X = 29 - 5v -56u + 176v - 16 - u = 13 + 171v - 57u. (13)
16--Substituting (11), (12), and (13) into (1)---> W + 13 + 171v - 57u + 5v + 28u - 88v + 8 = 100. (14)
17--Simplifying (14)---> W = 29u - 88v + 79. (14)
18--From (11) v =/> 1 and from (12) u =/> 3.
19--Trying v = 1 and u = 3, W = 78, X = 13, Y = 5, and Z = 4.
20--Checking--->78 + 13 + 5 + 4 = 100. Okay.
21--Checking--->78(.59) + 13(1.99) + 5(2.87) + 4(3.44) = 46.02 + 25.87 + 14.35 + 13.76 = 100. Okay.
22--Trying v = 2 and u = 3, X = 184 exceeding total of 100, therefore invalid.
23--Trying v = 1 and u = 4, X = negative number, therefore invalid.
24--All other values of u and v produce invalid results.
25--Therefore W = 78, X = 13, Y = 5, and Z = 4 is the only solution.

This is no doubt more than you asked for but I thought you you might be interested in the fact that N equations with less than N variables can be solved.

Denis
07-07-2006, 05:00 PM
These are therefore the only six combinations that will result in integer answers. Going through the arithmetic, you will see that all the combinations all add up to \$2.43.

True, BUT you DO need a 2nd equation to get a unique solution.

TchrWill
07-07-2006, 07:22 PM
These are therefore the only six combinations that will result in integer answers. Going through the arithmetic, you will see that all the combinations all add up to \$2.43.

True, BUT you DO need a 2nd equation to get a unique solution.

A second equation in terms of x and y is not required.