System of linear equations - solve with 3 variables?

[k9handler]

Fellas,
Here is an example problem, whats the best way to solve for X, Y and Z ?

First one I did come up undefined after two pages.


1.

X + 10Y + Z = 52
5X + Y + 4Z = 15
X + 2Y - 3Z = 12

Thanks in advance for any help you can provide.

[island911]

'best' depends on whether the eq's are from an algebra class, a pre-calc class, or a linear algebra class.

[k9handler]

Well this is from a basic college 101 algebra class.

[k9handler]

Found this info on a site.

* Solve the following system of equations using Gaussian elimination.

–3x + 2y – 6z = 6
5x + 7y – 5z = 6
x + 4y – 2z = 8

No equation is solved for a variable, so I'll have to do the multiplication-and-addition thing to simplify this system. In order to keep track of my work, I'll write down each step as I go. But I'll do my computations on scratch paper. Here is how I did it:

The first thing to do is to get rid of the leading x-terms in two of the rows. For now, I'll just look at which rows will be easy to clear out; I can switch rows later to get the system into upper triangular form. There is no rule that says I have to use the x-term from the first row, and, in this case, I think it will be simpler to use the x-term from the third row. So I'll multiply the third row by 3, and add it to the first row. I do the computations on scratch paper:

[-3x + 2y - 6z = 6] + [3x + 12y - 6z = 24] = [14y - 12z = 30]

...and then I write down the results:

system of equations with updated first row

(When we were solving two-variable systems, we could multiply a row, rewriting the system off to the side, and then add down. There is no space for this in a three-variable system, which is why we need the scratch paper.)

Since I didn't actually do anything to the third row, I copied it down, unchanged, into the new matrix. I used the third row, but I didn't actually change it. Don't let this confuse you.

To get smaller numbers, I'll multiply the first row by one-half:

first row: 7y - 6z = 15

Now I'll multiply the third row by –5 and add this to the second row. I do my work on scratch paper:

[5x + 7y - 5z = 6] + [-5x - 20y + 10z = -40] = [-13y + 5z = -34]

...and then I write down the results: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

system of equations with updated second row

I didn't do anything with the first row, so I copied it down unchanged. I worked with the third row, but I only worked on the second row, so the second row is updated and the third row is copied over unchanged.

Okay, now the x-column is cleared out except for the leading term in the third row. So now I have to work on the y-column. If I add twice the first row to the second this will give me a leading 1 in the second row. I won't have gotten rid of the leading y-term in the second row, but I will have converted it (without getting involved in fractions) to a form that is simpler to deal with. (You should keep an eye out for this sort of simplification.) First I do the scratch work:

[-13y + 5z = -34] + [14y - 12z = 30] = [y - 7z = -4]

...and then I write down the results:

system of equations with updated second row

Now I can use the second row to clear out the y-term in the first row. I'll multiply the second row by –7 and add. First I do the scratch work:

[7y - 6z = 15] + [-7y + 49z = 28] = [43z = 43]

...and then I write down the results:

system of equations with updated first row

I can tell what z is now, but, just to be thorough, I'll divide the first row by 43. Then I'll rearrange the rows to put them in upper-triangular form:

x + 4y - 2z = 8; y - 7z = -4; z = 1

Now I can start the process of back-solving:

y – 7(1) = –4
y – 7 = –4
y = 3

x + 4(3) – 2(1) = 8
x + 12 – 2 = 8
x + 10 = 8
x = –2

Then the solution is (x, y, z) = (–2, 3, 1).

and the above has me confused!!!

[m21sniper]

My first day ever in 9th grade algebra i showed up tripping face on LSD.

I never recovered.

[island911]

It's been a while since I've done it the hard way, but I think you want to start with rewriting the eq's in terms of the var's..

X + 10Y + Z = 52
5X + Y + 4Z = 15
X + 2Y - 3Z = 12

become

Z = 52-X -10Y
Y= 15 - 4Z - 5X
X = 12 - 2Y - 3Z

plug and chug.

[jyl]

Originally Posted by k9handler

Fellas,
Here is an example problem, whats the best way to solve for X, Y and Z ?

First one I did come up undefined after two pages.


1.

X + 10Y + Z = 52
5X + Y + 4Z = 15
X + 2Y - 3Z = 12

Thanks in advance for any help you can provide.

(1) X + 10Y + Z = 52
(2) 5X + Y + 4Z = 15
(3) X + 2Y - 3Z = 12


From (1)
(4) X = 52 - 10Y - Z
Substitute (4) into (2)
5 * (52 - 10Y - Z) + Y + 4Z = 15
260 - 50Y - 5Z +Y + 4Z = 15
49Y + Z = 245
(5) Z = 245 - 49Y
Substitute (4) and (5) into (3)
52 - 10Y - Z + 2Y - 3Z = 12
8Y + 4Z = 40
8Y + 4 * (245 - 49Y) = 40
8Y + 980 - 196Y = 40
188Y = 940
Y = 940 / 188
(6) Y = 5
Use (6) into (5) to solve Z
Z = 245 - 49 * 5
(7) Z = 0
Use (6) and (7) into (4) to solve X
X = 52 - 10 * 5 - 0
(8) X =2

Check the math please, as I didn't

Edit: OK, I'll finish the example.

To check, no need to go through all the steps to reverify, simply take X =2, Y = 5, and Z = 0 and plug into (1), (2), and (3) to see if all three equations come out right.
(1) X + 10Y + Z = 52
becomes
2 + 10 * 5 + 0 = 52, yup
(2) 5X + Y + 4Z = 15
becomes
5 * 2 + 5 + 4 * 0 = 15, yup
(3) X + 2Y - 3Z = 12
becomes
2 + 2 * 5 - 3 * 0 = 12, yup

The method on the website you found seems kind of complicated, not that it is wrong but makes too big a deal of things. Basically you grab one three-variable equation, re-arrange until one variable is expressed as a function of the other two. Plug the re-arranged equation into the next three-variable equation and get a two-variable equation, re-arrange until one variable is expressed as a function of the last variable. Plug both re-arranged equations into the last three-variable equation and you will have a one-variable equation, so that last variable is now solved to a constant. Then work backwards to solve the other variables.

[k9handler]

John.....thanks. My daughter was working it another way and came up with the same answers and it all checks out. It did help to see it done your way.

again, thank you.

[BlueSkyJaunte]

Originally Posted by island911

plug and chug.

I've always loved that phrase, sounds like a great name for a frat party.

[k9handler]

OK the last one...it has me and my daughter stuck.

31X + 12Y - 24Z = 105
18X + 24Y + z = 77
12X - 6Y + 4Z = 5

[speedracing944]

plug into Maple and have it do the hard work for you.

You could always turn it into a 3X4 matrix and solve it that way. Much easier.

Speedy

[k9handler]

Originally Posted by speedracing944

plug into Maple and have it do the hard work for you.

You could always turn it into a 3X4 matrix and solve it that way. Much easier.

Speedy

uh you lost me...maple? 3x4? Yeah I am that dumb.

[speedracing944]

Maple is a pretty awsome equation solving program but it cost big $$$ to purchase. for the matrix you would set it up like

31 12 -34 105
18 24 1 77
12 -6 4 5

Reduce it to reduced row echelon form and back substitute.

Speedy

[k9handler]

ok I found the maplesoft site, not sure I would want to purchase software just for this class.
The matrix one you eliminated the variables but how is it solved?

[speedracing944]

if you have a TI83 it will solve this stuff easy as pei

Speedy

[k9handler]

we are using TI82 but not sure how to use it for this.

BTW...your LT-1 944S is going to be crazy, amazed at how it fit.

[speedracing944]

I am not familiar with the TI82 but does it have a key marked "matrix"?

If it does then you should be able to enter in the matrix. Once it is entered you can solve the matrix using the RREF function which stands for reduced row echelon form. you will get a matrix which looks like..
1 0 0 number
0 1 0 number
0 0 1 number

the "number" in the first row would be your answer for X
the "number" in the second row would be Y
the last one would be you answer for Z

Speedy

[k9handler]

Originally Posted by speedracing944

I am not familiar with the TI82 but does it have a key marked "matrix"?

If it does then you should be able to enter in the matrix. Once it is entered you can solve the matrix using the RREF function which stands for reduced row echelon form. you will get a matrix which looks like..
1 0 0 number
0 1 0 number
0 0 1 number

the "number" in the first row would be your answer for X
the "number" in the second row would be Y
the last one would be you answer for Z

Speedy

it does and we entered all the numbers in a 3x4 matrix, but can't find the RREF function?

[speedracing944]

try this. look down at #6 on the link. X_1 would be your X, X_2 would be your Y and X_3 would be your Z
http://wwwstaff.murdoch.edu.au/~kissane/TI%20sheets/TI-82%20matrixeqns.pdf

Speedy

[speedracing944]

How to Solve Simultaneous Linear Equations Using Matrices on the TI-82

1) Arrange the linear equations to be solved so they have the form Ax + By = C.
Example: Using the equations above, we have 3x + 2y = 5 and 5x +4y = 27.

2) Write these equations in matrix form, using the coefficients A, B, and C to fill in the matrix.
The matrix for these equations would look like this:
3 2 5
5 4 27
The first column is a list of the x coefficients, the second column is a list of the y coefficients, and the third column is a list of values.

3) Turn on the TI-82, hit the [Matrix] key, select "Edit", select the name of a matrix that you want to use to solve this problem (A-E), and hit [Enter].
Example: We'll say that you've chosen to use matrix A for the coefficient matrix.

4) Define the size of your coefficient matrix.
Example: In this case, our coefficient matrix is two rows-by-two columns, so press "2" [Enter] "2" [Enter]. A problem that requires solving three equations in three unknowns will require a 3x3 coefficient matrix.

5) Enter the coefficients into the matrix in the calculator and then Quit.
Example: Press "3" [Enter] "2" [Enter] "5" [Enter] "4" [Enter][Quit]

6) Hit the [Matrix] key, select "Edit", select the name of a second matrix that you want to use to solve this problem (A-E), and hit [Enter].
Example: We'll say that you've chosen to use matrix B for the value matrix.

7) Define the size of your value matrix.
Example: In this case, our value matrix is two rows-by-one column, so press "2" [Enter] "1" [Enter]. A problem that requires solving three equations in three unknowns will require a 3x1 coefficient matrix.

8) Enter the values into the matrix in the calculator and then Quit.
Example: Press "5" [Enter] "27" [Enter][Quit]

9) Hit the [Matrix] key again, select Matrix A, hit [Enter], hit [x^-1], hit[*], hit [Matrix], select Matrix B, hit [Enter], then hit [Enter] again.
Example: [A]^-1 * [B]
NOTE: You MUST select the matrix name from the Matrix menu -- you cannot simply type in [A]!
This command tells the calculator to invert the first matrix, then multiply it by the second matrix, to get a matrix that lists your solutions in order. It's not critical that you understand this step, although the mathematical process is pretty cool!

10) Interpret Your Answer.
What appears on the calculator screen is this:
[[-17]
28]]
The -17 is the solution for the first variable, "x", and the 28 is the solution for the second variable, "y".



Using matrices to help you find solutions to simultaneous linear equations is one of the ways that graphing calculators can assist your problem-solving in physics. Please make sure, however, that you don't become TOO dependent on the calculator. Being able to manipulate variables and equations "by hand", as well as being able to substitute and solve equations in several unknowns is a valuable skill that cannot be replaced with a calculator, however powerful it may be.

http://www.ecs.umass.edu/~chang/TI83.htm

Speedy