System of linear equations question

[Ultima]

edit:
Solved.

Thanks for your help guys, and I got a response back from my teacher.

Here is what he says:

Hi Kevin, all you step are correct and from the last matrix you can
draw the conlusion, namely this system is always solvable for any b.
In fact you can read off the solution y=(b-4)/12 etc. ITI

Orig. Message

I have a test tomorrow and I'm not sure how to solve these types of problems. Someone please help

For which values of b the following linear system is consistent?

x - 3y = 2
2x + 6y = b
2x - 6y = 4

I can get the 12y = b - 4 from reducing the matrix, as I end up with:

[ 1 -3 2 ]
[ 0 12 b-4]
[ 0 0 0]

But how does this tell me which values of b this system is consistent. That's what I forget.

 

[sygyzy]

Isolate a variable. Say x. So you have x = 2 + 3y

So you have 2x + 6y = b which is 2 (2+3y) + 6y = b

or 4 + 6y + 6y = b or 4 +12y = b

12y = b - 4

y = (b-4)/12

EDIT: yes I know I transposed some numbers earlier.

 

[yankeesfan]

what grade are you in?

 

[ActuaryTm]

Originally posted by: Ultima
I have a test tomorrow and I'm not sure how to solve these types of problems.

Hopefully you're still within the withdrawal period.

 

[ed21x]

Originally posted by: sygyzy
Isolate a variable. Say x. So you have x = 2 + 3y

So you have 2x + 6y = b which is 2 (2+3y) + 6y = b

or 4 + 6y + 6y = b or 4 +12y = b

12y = b - 4

y = (b-4)/12

EDIT: yes I know I transposed some numbers earlier.

in Liniear Algebra, there is a different format to accomplish this. I'm pretty sure it involves putting them in a matrix, reducing the matrix, and then shuffling things around. This definitely isn't a regular algebra prob.

 

[imported_vr6]

Originally posted by: ed21x

Originally posted by: sygyzy
Isolate a variable. Say x. So you have x = 2 + 3y

So you have 2x + 6y = b which is 2 (2+3y) + 6y = b

or 4 + 6y + 6y = b or 4 +12y = b

12y = b - 4

y = (b-4)/12

EDIT: yes I know I transposed some numbers earlier.

in Liniear Algebra, there is a different format to accomplish this. I'm pretty sure it involves putting them in a matrix, reducing the matrix, and then shuffling things around. This definitely isn't a regular algebra prob.

yeah i am sure your professor doesn't want a simple subsituttion, u have to put it into a matrix and do row reductions

 

[Ultima]

Originally posted by: ed21x

Originally posted by: sygyzy
Isolate a variable. Say x. So you have x = 2 + 3y

So you have 2x + 6y = b which is 2 (2+3y) + 6y = b

or 4 + 6y + 6y = b or 4 +12y = b

12y = b - 4

y = (b-4)/12

EDIT: yes I know I transposed some numbers earlier.

in Liniear Algebra, there is a different format to accomplish this. I'm pretty sure it involves putting them in a matrix, reducing the matrix, and then shuffling things around. This definitely isn't a regular algebra prob.

Right. I can get the 12y = b - 4 from reducing the matrix, as I end up with:

[ 1 -3 2 ]
[ 0 12 b-4]
[ 0 0 0]

But how does this tell me which values of b this system is consistent. That's what I forget.

 

[sygyzy]

Originally posted by: ed21x

Originally posted by: sygyzy
Isolate a variable. Say x. So you have x = 2 + 3y

So you have 2x + 6y = b which is 2 (2+3y) + 6y = b

or 4 + 6y + 6y = b or 4 +12y = b

12y = b - 4

y = (b-4)/12

EDIT: yes I know I transposed some numbers earlier.

in Liniear Algebra, there is a different format to accomplish this. I'm pretty sure it involves putting them in a matrix, reducing the matrix, and then shuffling things around. This definitely isn't a regular algebra prob.

I took HS math, I swear. I forgot about Matrices. I know they exist and what they look like but forgot how to do it.

 

[Ultima]

^ going to sleep now.. gotta wake up early. Hopefully someone out there knows..

 

[TuxDave]

Originally posted by: Ultima
I have a test tomorrow and I'm not sure how to solve these types of problems. Someone please help

For which values of b the following linear system is consistent?

x - 3y = 2
2x + 6y = b
2x - 6y = 4

I can get the 12y = b - 4 from reducing the matrix, as I end up with:

[ 1 -3 2 ]
[ 0 12 b-4]
[ 0 0 0]

But how does this tell me which values of b this system is consistent. That's what I forget.

Dump it in a matrix. Solve B such that the determinent is.... 0? Btw.. I believe you screwed something up in your above matrix example.

 

[Howard]

??? The constant B somehow became a coefficient?

 

[Ultima]

comeon guys, my test is today. Nobody knows???

 

[Orsorum]

x - 3y = 2
2x + 6y = b
2x - 6y = 4

[1 -3] [2]
[2 6] [B ]
[2 -6] [4]
==>
[2 -6] [4]
[2 6]
[0 0] [0]
==>
[2 -6] [4]
[4 0] [b+4]
[0 0] [0]
==>
[2 0] [b/2+2]
[0 -6] [2-b/2]
[0 0] [0]
==>
[1 0] [b/4+1]
[0 1] [-1/3+b/12]
[0 0] [0]

 

[TuxDave]

Originally posted by: Ultima
I have a test tomorrow and I'm not sure how to solve these types of problems. Someone please help

For which values of b the following linear system is consistent?

x - 3y = 2
2x + 6y = b
2x - 6y = 4

I can get the 12y = b - 4 from reducing the matrix, as I end up with:

[ 1 -3 2 ]
[ 0 12 b-4]
[ 0 0 0]

But how does this tell me which values of b this system is consistent. That's what I forget.

FAWK... nm. You row reduced right. What's the definition of consistant? Because b could be anything and you could find a corresponding x and y.

 

[Goosemaster]

Originally posted by: Orsorum
x - 3y = 2
2x + 6y = b
2x - 6y = 4

[1 -3] [2]
[2 6] [B ]
[2 -6] [4]
==>
[2 -6] [4]
[2 6]
[0 0] [0]
==>
[2 -6] [4]
[4 0] [b+4]
[0 0] [0]
==>
[2 0] [b/2+2]
[0 -6] [2-b/2]
[0 0] [0]
==>
[1 0] [b/4+1]
[0 1] [-1/3+b/12]
[0 0] [0]

aye..finally..someone who knows what he is talking about

Row reduced to get to reduced echelon....solve....answer..easy...read your damn book

 

[Ultima]

Thanks for your help guys, and I got a response back from my teacher.

Here is what he says:

Hi Kevin, all you step are correct and from the last matrix you can
draw the conlusion, namely this system is always solvable for any b.
In fact you can read off the solution y=(b-4)/12 etc. ITI