# Solving a System of Three Linear Equations Using Elimination

#### tristatefabricatorsinc

##### Junior Member
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

##### Moderator
Staff member
Right up front. => -2x - 4z = 0

Be VERY careful.

#### stapel

##### Super Moderator
Staff member
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

##### Full Member
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
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

##### Moderator
Staff member
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

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

#### stapel

##### Super Moderator
Staff member
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.