Solving a System of Three Linear Equations Using Elimination

[tristatefabricatorsinc]

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]

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

Be VERY careful.

 

[stapel]

Re: Solving a System of Three Linear Equations Using Elimina

-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]

Re: Solving a System of Three Linear Equations Using Elimina

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]

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]

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]

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]

Are you supposed to subtract (2) from (1) or add them? I thought you were supposed to add them.

 

[stapel]

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, -1R₁ to R₂ and subtracting R₁ from R₂: -1R₁ + R₂ = R₂ - R₁.

Eliz.