• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

System of linear equations question

U

Ultima

Platinum Member
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.

 
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.
 
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.
Click to expand...

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.
 
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.
Click to expand...

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.
Click to expand...

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

 
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.
Click to expand...

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.
Click to expand...

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.
 
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.
Click to expand...

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.
Click to expand...



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. 🙁
 
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.
Click to expand...

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.
 
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]
 
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.
Click to expand...

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.
 
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]
Click to expand...


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

Row reduced to get to reduced echelon....solve....answer..easy...read your damn book😛
 
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
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top