system of linear equations  TOPIC_SOLVED

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system of linear equations

Postby Luke53 on Thu Apr 07, 2011 7:57 am

Is there an easy way to solve a system like this one?
x y/(3x-4y)=2/11
y z/(2y+3z)=6/5
x z/(x-z)=3/2
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Postby stapel_eliz on Thu Apr 07, 2011 1:28 pm

Luke53 wrote:x y/(3x-4y)=2/11
y z/(2y+3z)=6/5
x z/(x-z)=3/2

Since this is not a linear system, any solution method will likely be messy.

What have you tried so far? :wink:
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Re: system of linear equations

Postby Luke53 on Thu Apr 07, 2011 4:48 pm

Giving another example, of how it should be done according to my textbook:
given the following system:
x y/(x+y)=8/3
y z/(y+z)=8/5
x z/(x+z)=4/3
the first eqn is the same as:
(1/x)+(1/y)=3/8 (1)
the second eqn is:
(1/y)+(1/z)=5/8 (2)
the third eqn:
(1/x)+(1/z)=6/8 (3)
So the sum of the three last equations is equal to: 2/x+2/y+2/z=14/8; dividing this by 2 gives:
1/x+1/y+1/z=7/8 (4)
given the eqn's (2) and (4): 1/x= (7/8)-(5/8)= 2/8 and so x=4
out of (3) and (4): 1/y=7/8-6/8= 1/8 so y=8
and at last 1/z= (7/8)-(3/8)=4/8 and z=2 (out of (1 and (4))
I can't see how this can be applied to the eqn's given in the first system that I asked for an easier solution (and doesn't seem to be a linear system).
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Re: system of linear equations  TOPIC_SOLVED

Postby nona.m.nona on Thu Apr 07, 2011 9:37 pm

Luke53 wrote:the first eqn is the same as:
(1/x)+(1/y)=3/8 (1)
the second eqn is:
(1/y)+(1/z)=5/8 (2)
the third eqn:
(1/x)+(1/z)=6/8 (3)

These are the same, assuming that none of x, y, and z is zero.

For convenience, rename so 8/x = X, 8/y = Y, and 8/z = Z. This gives you the following:



This is a linear system which can be solved. Then you will need to back-solve to find the values of x, y, and z.
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Re: system of linear equations

Postby Luke53 on Fri Apr 08, 2011 8:32 am

Is there an easy way to solve a system like this one?
x y/(3x-4y)=2/11
y z/(2y+3z)=6/5
x z/(x-z)=3/2
Luke
Can one transform the three equations into the 1/x + 1/y form like in the previous example?
Applying the tranformation to the third eqn: x z/(x-z) = 3/2, would be something like: (1/z) - (1/x) = 2/3
How about transforming the first two equations into this form?

Greetings;
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Re: system of linear equations

Postby nona.m.nona on Fri Apr 08, 2011 11:07 am

Luke53 wrote:Can one transform the three equations into the 1/x + 1/y form like in the previous example?

The same process would likely work.
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Re: system of linear equations

Postby Luke53 on Fri Apr 08, 2011 3:03 pm

I don't know how to do this so please help.
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Postby stapel_eliz on Fri Apr 08, 2011 6:04 pm

Luke53 wrote:I don't know how to do this so please help.

What part of the previous worked example do you not understand?

Please be specific. Thank you! :wink:
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Re: system of linear equations

Postby Luke53 on Sat Apr 09, 2011 1:34 pm

I know that : x y /(x+y) = 8/3 is the same as 1/x + 1/y = 3/8
Could this sort of transformation also be done with: x y / (3x - 4y) = 2/11 ? ( guess not, but I'm not sure).
Thanks.
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Postby stapel_eliz on Sat Apr 09, 2011 4:23 pm

Luke53 wrote:I know that : x y /(x+y) = 8/3 is the same as 1/x + 1/y = 3/8
Could this sort of transformation also be done with: x y / (3x - 4y) = 2/11 ?

As mentioned previously, "The same process would likely work." That means that the same sort of transformation likely can also be done with the posted system. So try it, and see what happens! :wink:
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