- #1

gladius999

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Thanks

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- Thread starter gladius999
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- #1

gladius999

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Thanks

- #2

Office_Shredder

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- #3

gladius999

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a+2b+c=d

2a-b-4c=d

a-c=d

u can assume d=1?

2a-b-4c=d

a-c=d

u can assume d=1?

- #4

Office_Shredder

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Ok, good example. You actually have to break this up into two cases:

1) d=0. If d=0, you can just solve the equation since there are three equations and three unknowns.

2) d =/= 0. If d is non-zero, divide both sides of every equation by d. Call A=a/d, B=b/d, C=c/d. Then we get A+2B+C=1, 2A-B-4C=1, A-C=1 Then every solution of (A,B,C) corresponds to a set of solutions (Ad,Bd,Cd,d) where d is arbitrary. So we can essentially assume that d=0 or d=1 since we can derive all the solutions from this

1) d=0. If d=0, you can just solve the equation since there are three equations and three unknowns.

2) d =/= 0. If d is non-zero, divide both sides of every equation by d. Call A=a/d, B=b/d, C=c/d. Then we get A+2B+C=1, 2A-B-4C=1, A-C=1 Then every solution of (A,B,C) corresponds to a set of solutions (Ad,Bd,Cd,d) where d is arbitrary. So we can essentially assume that d=0 or d=1 since we can derive all the solutions from this

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- #5

gladius999

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but 2) i dont get how u can divide both sides of every equation by zero.

another gas example:

P1V=n1RT1

P2V=n1RT2

where V and R are constants

when solving equations can u assume V and R is 1?

- #6

maverick_starstrider

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Office_Shredder

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Yeah, that was just a mistype. I fixed my post to reflect what you should actually do

- #8

gladius999

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thnx for ur quick reply guys, i will try apply ur ideas into my thinking:)

- #9

HallsofIvy

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