Solving a System of Three Linear Equations Using Elimination

tristatefabricatorsinc

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Jan 31, 2006
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60
I am trying to solve the following equation using elimination

. . .2x + 4y + 6z = 2
. . .x . . .. . .+ 2z = 0
. . .2x + 3y – z = -5

I want to start off by getting rid of the x variable - starting with equation 2, I did the following...

. . .-2(x + 2z = 0) => -2x - 2z = 0

Take this and add it to equation 1:

. .. .2x + 4y +6z = 2
. . .-2x. . . . . -2z = 0
. . .-------------------
. . .. . . .4y + 4z = 2

Now I will do the same thing to equation 2 and add it to equation 3:

. . .-2(x + 2z = 0) => -2x -2z = 0

. .. .2x + 3y - z = -5
. . .-2x . .. . - 2z = 0
. . .-------------------
. . .. .. . .3y - 3z = -5

Now I take my two answers and combine them:

. . .4y + 4z = 2: multiply by 3
. . .3y - 3z = -5: multiply by -4

to get...

. . . .12y + 12z = 6
. . .-12y + 12z = 20
. . .-------------------
. . .. . .. . .24z = 26

I am confused. It is not coming out even. Am I doing it wrong? Any help would be greatly appreciated. Thank you!
 

tkhunny

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Right up front. => -2x - 4z = 0

Be VERY careful.
 

stapel

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Re: Solving a System of Three Linear Equations Using Elimina

tristatefabricatorsinc said:
-2(x + 2z = 0) => -2x - 2z = 0
-2(+2z) = -4z, not -2z.

By the way, not all systems will have whole-number answers. Don't assume an error on that basis alone.

Eliz.
 

jonboy

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Re: Solving a System of Three Linear Equations Using Elimina

tristatefabricatorsinc said:
I am trying to solve the following equation using elimination

2x + 4y + 6z = 2
x + 2z = 0
2x + 3y – z = -5

I want to start off by getting rid of the x variable - starting with equation 2, I did the following...

-2(x + 2z = 0) = -2x -2z = 0

take this and add it to equation 1
2x + 4y +6z = 2
-2x -2z = 0
-------------------
4y + 4z = 2

Now I will do the same thing to equation 2 and add it to equation 3
-2(x + 2z = 0) = -2x -2z = 0.........That should be -2x -4z = 0
So this should be:
2x + 3y -z = -5
-2x -4z = 0
-------------------
3y - 5z = -5

Now I take my two answers and combine them...
4y + 4z = 2 multiply by 3
3y - 5z = -5 multiply by -4
to get...

12y + 12z = 6
-12y +20z = 20
-------------------
32z = 26

So z=(32/26) -------> (16/13)

I Am Confused, it is not coming out even, am I doing it wrong?

Any help would be greatly appreciated...
 

tristatefabricatorsinc

Junior Member
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Jan 31, 2006
Messages
60
Now that z is an improper fraction, how do I continue? Do I plug it in and go? Are fractional answers ok or does it mean something in particular?


2x + 4y + 6z = 2
x + 2z = 0
2x + 3y – z = -5

I want to start off by getting rid of the x variable - starting with equation 2, I did the following...

-2(x + 2z = 0) = -2x -2z = 0

take this and add it to equation 1
2x + 4y +6z = 2
-2x -2z = 0
-------------------
4y + 4z = 2

Now I will do the same thing to equation 2 and add it to equation 3
-2(x + 2z = 0) = -2x -4z = 0.........That should be -2x -4z = 0
So this should be:
2x + 3y -z = -5
-2x -4z = 0
-------------------
3y - 5z = -5

Now I take my two answers and combine them...
4y + 4z = 2 multiply by 3
3y - 5z = -5 multiply by -4
to get...

12y + 12z = 6
-12y +20z = 20
-------------------
32z = 26

So z=(32/26) -------> (16/13)
 

tkhunny

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tristatefabricatorsinc said:
Now that z is an improper fraction
Why does that matter. Just go with it.

-2(x + 2z = 0) = -2x -2z = 0
You still didn't fix the first problem. Please read previous posts.
 

soroban

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Jan 28, 2005
Messages
5,588
Re: Solving a System of Three Linear Equations Using Elimina

Hello, tristatefabricatorsinc!

\(\displaystyle \begin{array}{ccc}2x\,+\,4y\,+\,6z \:= \:2 \\ \;x\;\;\;+\;\;\;2z \:= \:0 \\ 2x\,+\,3y\,-\,z \:=\:-5\end{array}\)

Multiply the second equation by 2:

. . \(\displaystyle \begin{array}{ccc}2x\,+\,4y\,+\,6z \:= \:2 \\ \;2x\;\;\;+\;\;\;4z \:= \:0\; \\ 2x\,+\,3y\,-\,z \:=\:-5\end{array}\;\begin{array}{cc}(1)\\(2)\\(3)\end{array}\)

Subtract (2) from (1): \(\displaystyle \;4y\,+\,2z\:=\:\:2\;\;(4)\)
Subtract (2) from (3):\(\displaystyle \;\,3y\,-\,5y\:=\:\)-\(\displaystyle 5\;\;(5)\)

Multiply (4) by 5: \(\displaystyle \;20y\,+\,10y\:=\:10\;\;\;(6)\)
Multiply (5) by 2: \(\displaystyle \;\:6y\,-\,10z\:=\:\)-\(\displaystyle 10\;\;(7)\)

Add (6) and (7): \(\displaystyle \;26y \:=\:0\;\;\Rightarrow\;\;\)y = 0

Substitute into (4); \(\displaystyle \;4(0)\,+\,2z\:=\:2\;\;\Rightarrow\;\;\)z = 1

Substitute into (2): \(\displaystyle \;2x\,+\,4(1)\:=\:0\;\;\Rightarrow\;\;\)x = -2


Therefore: \(\displaystyle \;\)(x,y,z) = (-2,0,1)

 

tristatefabricatorsinc

Junior Member
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Jan 31, 2006
Messages
60
Are you supposed to subtract (2) from (1) or add them? I thought you were supposed to add them.
 

stapel

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Messages
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tristatefabricatorsinc said:
Are you supposed to subtract (2) from (1) or add them? I thought you were supposed to add them.
There is no difference between adding, say, -1R<sub>1</sub> to R<sub>2</sub> and subtracting R<sub>1</sub> from R<sub>2</sub>: -1R<sub>1</sub> + R<sub>2</sub> = R<sub>2</sub> - R<sub>1</sub>.

Eliz.
 
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