Question

**MEDAL**
Can someone please help me? I'm confused with this question.

Solve the system by elimintation.

Answer 1

So combine like terms then?

Answer 2

@austinL
@JjJustin
@lizzylou169
@pitamar
@satellite73
Please help D:

Answer 3

where are you getting stuck?

Answer 4

@wilfley
I don't understand how to solve it :/

Answer 5

I would start with adding equations 1 and 3 together and adding equations 2 and 3 together.

Answer 6

For adding 1 and 3 together, I got x + 5y + 6z = 5. Am I on the right track?

Answer 7

I added what I got from 1 and 3 and put it with 2.
Now I have -2x - 5y + 6z = 2. Now what, @wilfley ?

Answer 8

We have,

\((1)~-2x+2y+3z=0\\
(2)~~~~~-2x-y+z=-3\\
(3)~~~~~~2x+3y+3z=5\)

What we would first want to do is add line (1) and (3) together.

\(-2x+2y+3z=0\\
\underline{-2x-y+z=-3}\\
~0x+y+4z=-3\)

Then you would need to eliminate the \(x\) term from (1) and (2).

Doing this, you should have a system of two equations, two unknowns. I believe you should be able to solve for that. Once you find \(y\) and \(z\), you can easily solve for \(x\).

Answer 9

Ooooh!

Answer 10

So now @austinL , do I solve 0x + y + 4z = -3?

Answer 11

Combining (1) and (2) to get rid of the \(x\) term we should get,

\(3y+2z=3\)

Then we can combine that with the one we found earlier,

\((1.1)~y+4z=-3\\
(2.1)~3y+2z=3\)

Then you can solve that for \(z\) and \(y\).

Answer 12

I have to go now, hopefully, you can get help finishing the rest of the problem! Take care!

Answer 13

Thanks, @austinL !

Answer 14

-2x+2y+3z=0-----(1

-2x-y+z=-3-------(2

2x+3y+3z=5------(3

In order to solve such equations i.e. equations having three variables we first need to convert them in two variable equations. So, to do this let add 1& 3 and then 2 & 3. this would give two equations in two variables: Let us try out.
Adding 1 & 3, we find:
-2x+2y+3z=0-----(1
2x+3y+3z=5------(3
__________________________
5y +6z =5 ------(4

Adding 2 & 3, we find:
-2x-y+z=-3-------(2
2x+3y+3z=5------(3
___________________________
2y + 4z =2
i.e. y +2z= 1 (dividing both sides by 2)
i.e. y= 1-2z ------(5

Now by substitution method from (4 and (5 we have

5(1-2z) +6z =5
i.e. 5- 10z +6z=5

i.e. 5- 4z =5
i.e. - 4z =5-5
i.e. - 4z =0
i.e. z =0

Now from eq. (5 we have

y= 1-2z = 1-2*0=1-0=1
i.e. y=1

Now from eq. (3 we have

2x+3*1+3*0=5
i.e. 2x+3+0=5
i.e. 2x+3=5
i.e. 2x=5-3
i.e. 2x=2
i.e. x=2/2
i.e. x=1
Hence required solutions of the given system of equations are
x=1, y=1 & z=0

@NobodySpecial

Answer 15

THANK YOU SO MUCH.

Answer 16

@NobodySpecial You are most welcome.

Answer 17

@dpasingh
Do you think you can help me with one more?