Question

Help with the solution to the system.

Answer 1

Can someone help me?

Answer 2

You get rid of a variable, "y", by subtracting the second and third equation. Then you can solve simultaneously with the first equation and the equation you get when you subtract the last two equations.

Answer 3

Could you show me the equation you get from the last two equations? If you still have trouble finding that particular equation, let me know. Here's something for you to start off with (I will label the equation number at the end of each equation just to tell you what equation I took from your question):
\[2ax+y=0 [2]\]
\[ax+y+z=0[3]\]
\[[2]-[3]\]
\[(2ax+y)-(ax+y+z)=(0)-(0)\]

Answer 4

@RFJ-86 Are you with me?

Answer 5

Yes, I just trying to work it out. :)

Answer 6

\[(2ax+y)-(ax+y+z)=0\]
And then find a?
\[a=z/x\]

Answer 7

Nope, I want you to simplify what I wrote down.

Answer 8

Hmm
\[ax-z=0\]

Answer 9

Yes. Now you have a new equation. We will label that equation as [4].

Answer 10

So now we use the first equation and this equation and solve simultaneously.

Answer 11

\[4x-az=0[1]\]
\[ax-z=0[4]\]
\[[1]\times a\]
\[(4x-az)\times a=0\times a\]
\[4ax-a^2z=0[5]\]
\[[2]\times 4\]
\[(ax-z)\times 4=0\times 4\]
\[4ax-4z=0[6]\]
\[[6]-[5]\]

Answer 12

Could you tell me what you get when you subtract equation 6 with equation 5?

Answer 13

\[(4ax-4z)-(2ax-a^2z)\]

Answer 14

\[=za^2+2xa-4z\]

Answer 15

No, equation 5 is 4ax-a^2z.

Answer 16

I deliberately made both equations have 4ax, so I can get rid of another variable, "x".

Answer 17

sorry..

Answer 18

z(a^2-4)

Answer 19

The trick to these questions is to eliminate one variable at a time.

Answer 20

Yes I see..

Answer 21

that would equal to zero right? So we can get rid of z by dividing z to both sides.

Answer 22

then you can solve for a. Remember when you deal with terms that are squaed, you will get two solutions.

Answer 23

squared*

Answer 24

2 and -2

Answer 25

if a=4 or -4, it has infinitely many solutions.
if a has any value other than 4, there is only one solution that is (0,0,0).

Answer 26

sorry..a=2 or -2.

Answer 27

Yep, correct.

Answer 28

Thank you @Azteck

Answer 29

No worries.