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Old March 12th, 06:23 PM   #1 (permalink)
Jinjiro
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Math Help (Pre-Calc)

x + y -5z = 3
x - 2z = 1
2x - y - z = 0

Solve using elimination.

If you try to eliminate y, you wind up with the same equation twice: 3x - 6z = 3

So I'm kind of stuck.

-Jin
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Old March 12th, 06:27 PM   #2 (permalink)
NYY
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3x-6z=3 is fine
then multiply the second equation by -3 to get -3x +6z = -3
add that to your 3x - 6z = 3 to get 0 = 0
That means that you have infinite solutions. So set one of your variables equal to "a", math teachers tell you to make it z. so if z=a, plug it in to your other equations, and get your x and y values in terms of a.
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Old March 12th, 06:37 PM   #3 (permalink)
TonyD
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I gave it a stab another way, too. I'm not sure it's solvable in the sense that you'll get numbers for x,y,z.
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Old March 12th, 06:47 PM   #4 (permalink)
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Right, there are infinite solutions to it.
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Old March 12th, 06:48 PM   #5 (permalink)
Jinjiro
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Are you supposed to make z equal to the equation 3x - 6z = 3?
So would it be like z = (1/2)x + -(1/2)?

And then would I just plug that in to z for the other equations and solve?

-Jin
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Gentlemen, I have had men watching you for a long time and I am convinced that you have used the funds of the bank to speculate in the breadstuffs of the country. When you won, you divided the profits amongst you, and when you lost, you charged it to the bank. You tell me that if I take the deposits from the bank and annul its charter, I shall ruin ten thousand families. That may be true, gentlemen, but that is your sin! Should I let you go on, you will ruin fifty thousand families, and that would be my sin! You are a den of vipers and thieves.
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Old March 12th, 06:49 PM   #6 (permalink)
TonyD
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Quote:
Originally Posted by Jinjiro
Are you supposed to make z equal to the equation 3x - 6z = 3?
So would it be like z = (1/2)x + -(1/2)?

And then would I just plug that in to z for the other equations and solve?

-Jin
Tried that...something like that. All I got was 3=3, 1=1, and 0. All the variables subtract out.
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Old March 12th, 06:51 PM   #7 (permalink)
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Yeah, that's what I got. I guess they're all infinite solutions? Or something...

-Jin
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Gentlemen, I have had men watching you for a long time and I am convinced that you have used the funds of the bank to speculate in the breadstuffs of the country. When you won, you divided the profits amongst you, and when you lost, you charged it to the bank. You tell me that if I take the deposits from the bank and annul its charter, I shall ruin ten thousand families. That may be true, gentlemen, but that is your sin! Should I let you go on, you will ruin fifty thousand families, and that would be my sin! You are a den of vipers and thieves.
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Old March 12th, 07:01 PM   #8 (permalink)
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use a matrix... i'll solve it later if you don't know how to do it that way.
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Old March 12th, 07:11 PM   #9 (permalink)
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[[1 1 -5] -1
[1 -2 0]
[2 -1 -1]]

then multiply by

[[3]
[1]
[0]]

the first matrix shown is multiplied by the inverse, because the original eq A (x,y,z)= B
then it goes to:
(x,y,z)=A^-1 * B

my calculator gave me {-1, -1, -1}

so perhaps infinate solutions is correct?
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Old March 12th, 07:13 PM   #10 (permalink)
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Pre-calc? We did that this year in Algebra2H
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Old March 12th, 08:10 PM   #11 (permalink)
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Originally Posted by big_balla89
Pre-calc? We did that this year in Algebra2H

yea i was in Algebra 2 last year and we did the same stuff. this year in trig/precalc all we did was stuff with sin/cos/tan
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Old March 12th, 08:25 PM   #12 (permalink)
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Quote:
Originally Posted by mynameisgeorgel
yea i was in Algebra 2 last year and we did the same stuff. this year in trig/precalc all we did was stuff with sin/cos/tan
Trig...

We are doing that this year, too. We're dealing with sec, csc, and cot as well.

I don't like it because its in the morning and by the time I'm fully awake, he's through with the lesson and giving out the assignment... Bah.
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Old March 12th, 08:54 PM   #13 (permalink)
Jinjiro
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Quote:
Originally Posted by big_balla89
Pre-calc? We did that this year in Algebra2H
It's supplementary with matrices, so you'll probably do it again. We did it in algebra as well.

-Jin
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Gentlemen, I have had men watching you for a long time and I am convinced that you have used the funds of the bank to speculate in the breadstuffs of the country. When you won, you divided the profits amongst you, and when you lost, you charged it to the bank. You tell me that if I take the deposits from the bank and annul its charter, I shall ruin ten thousand families. That may be true, gentlemen, but that is your sin! Should I let you go on, you will ruin fifty thousand families, and that would be my sin! You are a den of vipers and thieves.
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Old March 12th, 09:00 PM   #14 (permalink)
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Yeah...we're doing linear algebra in Differential Equations.
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Old March 12th, 09:07 PM   #15 (permalink)
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Quote:
Originally Posted by big_balla89
Trig...

We are doing that this year, too. We're dealing with sec, csc, and cot as well.

I don't like it because its in the morning and by the time I'm fully awake, he's through with the lesson and giving out the assignment... Bah.

yea i left those out sec csc cot
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Old March 13th, 06:59 AM   #16 (permalink)
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Quote:
Originally Posted by big_balla89
Pre-calc? We did that this year in Algebra2H
same here... if it wasnt done solved i woulda gave it a shot...im still getting hang of it...we jus had a test over this last friday.
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Old March 13th, 03:50 PM   #17 (permalink)
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When the equation is put into standard matrix form it becomes a singular matrix, meaning it's determinant is zero and also has no inverse... and if i remember correctly, there are NO solutions because when you solve a matrix algebraically, such as we have to do in this case, you are in turn taking the inverse when you solve, (inverse operations), and you all know you can not divide by zero, which you have to do to solve this, so it has no REAL solution.
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