# Simultaneous eqs methodology (4 vars)

1. Oct 5, 2009

### jjc

I am looking for someone to explain the methodology behind solving simultaneous equations, specifically of four variables. I have read through the numerous other threads about this, and I get most of it, but it always appears that there is some rule or step that I am missing and I am unable to come up with a good answer for my own problem. (This is also my first try at using LaTex. I tried to make it nicely formatted but didn't succeed everywhere. It didn't want to give me new lines inside the Tex tags using the double \. :( )

1. The problem statement, all variables and given/known data
It really starts out as a request to solve a matrices equation:

Determine w, x, y, z so that this equation holds:
$$\left (\begin{array}{cccc} 2&1&-1&7 \\6&8&0&3 \end{array} \right) \left (\begin{array}{cc} x & 2x\\ y & -y+z \\ x+w & w-2y+x \\ z & z \end{array} \right) = - \left (\begin{array}{cc} 45 & 46\\ 3 & 87 \end{array} \right)$$

2. Relevant equations
Getting the equations out isn't too difficult. For my own benefit, here is all my work:

A size is 2x4, B size is 4x2; they can by multiplied; resultant array is 2x2 (n=4 for summation)

eq1: $$c_{11} = \sum^{n=4}_{j=1} a_{1j}\cdot b_{j1} = 2x + y - x - w + 7z = \underline{x + y - w + 7z}$$
eq2: $$c_{12} = \sum^{n=4}_{j=1} a_{1j}\cdot b_{j2} = 2\cdot2x + 1(-y + z) + -1 (w-2y+x) + 7z = \underline{3x + y - w + 8z}$$
eq3: $$c_{21} = \sum^{n=4}_{j=1} a_{2j}\cdot b_{j1} = \underline{6x + 8y + 0 + 3z}$$
eq4: $$c_{22} = \sum^{n=4}_{j=1} a_{2j}\cdot b_{j2} = 12x -8y + 8z + 0 +3z = \underline{12x - 8y + 11z}$$

And each of the equations above equals the corresponding element from the result matrix.

3. The attempt at a solution

Now, I have run through these sets 4 or 5 times using different methods (elimination or substitution) but always get answers that don't check. Here was one of my more promising runs:

Eq1: $$x + y - w + 7z = -45 \\$$
Eq2: $$\underline{(-1) (3x + y - w + 8z) = -46 (-1)} \ \ \ \mbox {(multiply by -1 so can add and eliminate)} }$$
Eq A: $$\ \ \ \ -2x + 0 + 0 - z = 1 \ \ \ \ => \ \ \ \underline{z = -1 - 2x}$$

Now, substitute Eq A into Eq 3:

Eq B: $$6x + 8y + 3(-1 - 2x) = -3 \ \ \ \ => \ \ \ 8y - 3 = -3 \ \ \ \ => \ \ \ 8y = 0 \ \ \ \ => \ \ \underline{y = 0}$$

Now here is where I think things start going wrong. Can I take Eq A and Eq B and throw them back into Eq 4?

Eq C: $$12x - 8\cdot0 + 11 (-1 - 2x) = -87 \ \ \ => \ \ 12x - 11 - 22x = -87 => -10x = -76 \ \ \ \ => \ \ \ \underline { x = 7.6 }$$

OK, not looking too bad there, but I now have two values, and a couple of equations where I can plug values into to find the third: Eq 3, Eq 4, and Eq A. Do I have to throw those values back into original equations and not derived equations (e.g. 3 or 4, but not A)? Can I just put that back into A to find z, then use x, y, and z in any of the other ones to find w?

Eq 3 (with value replacements):
$$6(7.6) + 8(0) + 3z = -3 \ \ \ \ => \ \ \ \ 45.6 + 3z = -3 \ \ \ \ => \ \ \ z = \frac{-3 - 45.6}{3} = \underline{16.2}$$

And lastly, Eq 1 with value replacements:
$$7.6 + 0 - w + 7(16.2) = -45 \ \ \ \ => \ \ \ \underline {w = 76}$$

Hmmmm...this is my cleanest answer yet, and the values for z and w in the last two equations are different than what I have gotten before...

Eq2 : $$3(7.6) + 0 - 76 + 8(16.2) = -46 \\ 22.8 -76 + 129.6 = -46 \ \ \ \ \ \ ?$$

Uh, nope. See! I need to know the background rules so that I can do this correctly and repeatedly instead of haphazardly! And I probably made a number of minor algebraic or arithmetic errors in there, too. :)

Thanks for any pointers!

-- jc

Last edited by a moderator: Oct 5, 2009
2. Oct 5, 2009

### LCKurtz

I haven't checked your work. But I have a question about your original matrix multiplication. Your two matrices aren't conformable. You can't multiply a 4x2 matrix by a 4x2 matrix. You could multiply (2x4) times (4x2) to get a 2x2. So it may be that you haven't transcribed your problem correctly into matrix form in the first place.

3. Oct 5, 2009

### Integral

Staff Emeritus
I am with LC, You need to transpose one of those matrices to do the multiplication. I am not seeing how your rows and columns match up to get your equations. It that the starting point you are given or is there more behind those matrices?

OK, I have looked at your latex code, and found the problem. I changed the {cc} to {cccc} now it looks the way it should.

A good way to systematize solving things like this is to create what is called the augmented matrix from your 4 equations. Write each of your equations with the variables in the same order. Now take only the coefficients for each variable and create a 4x4 matrix with them, preserving the order imposed by the variables. Create the augmented column by adding a 5th column consisting of the constant (the elements of your 2x2 matrix) lined up with the appropriate equation.

you can now do matrix operations to obtain a 4x4 identity matrix the final column will be the value of the variable.

Check your Private messages (upper right side of the screen)

Last edited: Oct 5, 2009
4. Oct 9, 2009

### jjc

Thanks all for the suggestions. And thanks for correcting my Tex. I think I mentioned that it was my first time through it. :)

Sorry for not getting back to this until now. Things have been hectic with school. However, the methodology you suggest is beyond me! I did stop in at the math lab to get some help. He was telling me that there weren't necessarily any hard and fast rules about where to plug things in; any of the derived equations should be valid in any of the other equations. We worked through it and came up with the right answers; it seems I had a few things right, already. I was able to work through it again later and get the same answers (well, I did use the same pattern as he had showed me, too).

I essentially added eq1 and eq2 to get eqA. Then I added eq3 and eq4 to get eqB. EqA and eqB then only had two terms. Then I added eqA and eqB to get the value for one variable. Then I used that value back into eqA or eqB to get the second value. Then taking those two values and putting them back into eq3 or eq4 (the ones with only 3 variables) to get the third value, then taking all three back into eq1 or eq2 to get the fourth value.

It seemed so obvious when he was doing it! :)

-- jc

5. Oct 9, 2009

### HallsofIvy

Staff Emeritus
I would suggest that, instead, you add the third and fourth equations. That will eliminate y leaving you with two equations in x and z, 2z+ z= -1 and 18x+ 14z= -90. Dividing that last equation by 2, 9x+ 7y= -45. Now put z= -1-2x in that equation to get 9x+7(-1-2x)= -5x- 7= 45.