## system of linear equations

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.

### system of linear equations

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
Luke53

Posts: 54
Joined: Sun Mar 13, 2011 9:46 am

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?

stapel_eliz

Posts: 1793
Joined: Mon Dec 08, 2008 4:22 pm

### Re: system of linear equations

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).
Luke.
Luke53

Posts: 54
Joined: Sun Mar 13, 2011 9:46 am

### Re: system of linear equations

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:

$\begin{array}{cccccc}X&+&Y&\,&=&3\\\,&\,Y&+&Z&=&5\\X&&+&Z&=&6\end{array}$

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.
nona.m.nona

Posts: 256
Joined: Sun Dec 14, 2008 11:07 pm

### Re: system of linear equations

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;
Luke.
Luke53

Posts: 54
Joined: Sun Mar 13, 2011 9:46 am

### Re: system of linear equations

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.
nona.m.nona

Posts: 256
Joined: Sun Dec 14, 2008 11:07 pm

### Re: system of linear equations

Luke.
Luke53

Posts: 54
Joined: Sun Mar 13, 2011 9:46 am

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

stapel_eliz

Posts: 1793
Joined: Mon Dec 08, 2008 4:22 pm

### Re: system of linear equations

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.
Luke53

Posts: 54
Joined: Sun Mar 13, 2011 9:46 am

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!

stapel_eliz

Posts: 1793
Joined: Mon Dec 08, 2008 4:22 pm

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