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 March 9th, 2017, 04:30 PM #1 Member   Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 Impossible simultaneous? So I've got this issue. The main issue is an electrical one but at this point it's just math, so I've put it here in this forum. I've ended up with these three equations. $\displaystyle 4 = 20_{i1} - 15_{i2}$ eq1. $\displaystyle 0 = 15_{i1} + 33_{i2} - 8_{i3}$ eq2. $\displaystyle -6 = -8_{i2} + 20 _{i3}$ eq3. At this point I usually use simultaneous equations to solve the variables. I don't think I can with these one though. Usually I substitute in a zero value like for eq1: $\displaystyle 4 = 20_{i1} - 15_{i2}$ becomes $\displaystyle 4 = 20_{i1} - 15_{i2} + 0_{i3}$ This time I have to substitute a zero value for Eq3. as well. $\displaystyle -6 = 0_{i1} -8_{i2} + 20 _{i3}$ But here's the issue - then you can't actually multiply zero by anything Neither can you divide a number by anything to actually arrive at zero. So therefore I can't cancel out to solve - if you see what I mean??? Last edited by Kevineamon; March 9th, 2017 at 04:34 PM.
 March 9th, 2017, 04:40 PM #2 Member   Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 Just looking at it now - silly me. If I just sub in for eq3. cancel out the $\displaystyle _{i3}$ then solve... Is that right?? hmmm...
March 9th, 2017, 06:10 PM   #3
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Quote:
 Originally Posted by Kevineamon Just looking at it now - silly me. If I just sub in for eq3. cancel out the $\displaystyle _{i3}$ then solve... Is that right?? hmmm...
What does the subscript i1 mean? Are these currents in a circuit?

-Dan

 March 9th, 2017, 11:41 PM #4 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Do you know how to set it up as an Augmented Matrix? If you do then Row Reduced Echelon Form will give you all the solutions at once. Thanks from Kevineamon
 March 10th, 2017, 12:48 AM #5 Member   Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 I must learn how to do that Agent. I've seen people do it. Never tried. Yeh Dan these are currents. But as I said the problems simply math at this point. If we said they were XYZ it would be all the same. Hence I posted here.
March 10th, 2017, 04:51 AM   #6
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Quote:
 Originally Posted by Kevineamon . . . you can't actually multiply zero by anything
Yes, you can. The result is zero. The equations you gave have a unique solution that can be found quite easily.

 March 10th, 2017, 06:02 AM #7 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Since the name of the variable doesn't matter ... $\ \ \ \ 4 = 20x \ - 15y$ $\ \ \ \ 0 = 15x + 33y \ - \ 8z$ $- \ 6 = \ \ \ \ \ \ \ \ - 8y + 20z$ I tried to line up the variable columns the best I could so the LaTex code would show better. Use eq2 and eq3 to eliminate $z$. Multiply eq2 by $5$ , eq3 by $2$ then add them. $\ \ \ \ \ \ 0 = 75x +165y - 40z$ $-12 = \ \ \ \ \ \ \ - \ \ 16y + 40z$ Adding these $2$ equations gives eq4 $-12 = 75x + 149y$ Now use eq1 and eq4 to eliminate $x$ , as you can see ... this is turning into a mess. I leave it to you.
 March 10th, 2017, 07:52 AM #8 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Using a CAS (computer algebra system) I get x = 416/4105, y = -108/821 and z = -579/1642.

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