Solving a system of equations .

[agoogler]

Homework Statement

Solve simultaneously:
6(x + y) = 5xy,
21(y + z) = 10yz,
14(z + x) = 9zx

Homework Equations

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The Attempt at a Solution

Obviously one of the solution is (0,0,0) But I'm more interested in finding the other.
Expanding these three equations , I get -
6x+6y=5xy
21y+21z=10yz
14z+14x=9zx

But what now? I've no experience solving these type of systems . Please give hints.

 

[pasmith]

Multiply each equation by the unknown which doesn't appear in it so that the right hand side of each equation is then a multiple of xyz. You can then eliminate xyz in three ways to end up with three linear simultaneous equations for xy, xz and yz.

Having solved those, use the fact that (xy)(xz)(yz) = (xyz)^2 to find x, y and z up to a sign.

 

[Ray Vickson]

Multiply each equation by the unknown which doesn't appear in it so that the right hand side of each equation is then a multiple of xyz. You can then eliminate xyz in three ways to end up with three linear simultaneous equations for xy, xz and yz.

Having solved those, use the fact that (xy)(xz)(yz) = (xyz)^2 to find x, y and z up to a sign.

Even better: if p = xyz, one can express xy, xz and yz as numerical multiples of p (by solving linear equations with p in the right-hand-sides), so---if x, y and z are non-zero---one gets a unique solution for (x,y,z), with no sign ambiguities.

Also: explore what must happen if one of the variables, say x, is zero.

 

[agoogler]

Thanks for the hints , but I used a different method to solve this equation. From eq. 1 I got y=(6x)/(5x-6) .
I substituted that in eq. 2 to get x in terms of z. ( x=(126z)/(126+45z) ) then I substituted that into eq. 3 to get a quadratic equation in z with solutions z=0,7. From z=7 , I got x= 2 . Then I got y=3 from the relationship between x and y obtained earlier from eq. 1 . But I felt that the arithmetic involved was quite daunting and there must be a better way to solve it. I tried using the method that pasmith recommended but am feeling a bit lost. Can anyone please explain a simpler method to solve this system ?

 

[Ray Vickson]

Thanks for the hints , but I used a different method to solve this equation. From eq. 1 I got y=(6x)/(5x-6) .
I substituted that in eq. 2 to get x in terms of z. ( x=(126z)/(126+45z) ) then I substituted that into eq. 3 to get a quadratic equation in z with solutions z=0,7. From z=7 , I got x= 2 . Then I got y=3 from the relationship between x and y obtained earlier from eq. 1 . But I felt that the arithmetic involved was quite daunting and there must be a better way to solve it. I tried using the method that pasmith recommended but am feeling a bit lost. Can anyone please explain a simpler method to solve this system ?

Let $a = xy, \;b = yz, \; c = zx \:\text{ and }\: p = xyz.$ Multiply the first equation by z, the second by x and the third by y to get
6b+6c=5p\\<br /> 21a+21c=10p \\<br /> 14a+14b=9p
This is a simple 3x3 linear system which can be solved using grade-school methods, to get
$a = p/7,\: b = p/2, \: c = p/3.$ Assuming $x,y,z,\neq 0$ we have
xy = xyz/7 \Longrightarrow z = 7 \\<br /> yz = xyz/2 \Longrightarrow x = 2\\<br /> zx = xyz/3 \Longrightarrow y=3

 

[agoogler]

Let $a = xy, \;b = yz, \; c = zx \:\text{ and }\: p = xyz.$ Multiply the first equation by z, the second by x and the third by y to get
6b+6c=5p\\<br /> 21a+21c=10p \\<br /> 14a+14b=9p
This is a simple 3x3 linear system which can be solved using grade-school methods, to get
$a = p/7,\: b = p/2, \: c = p/3.$ Assuming $x,y,z,\neq 0$ we have
xy = xyz/7 \Longrightarrow z = 7 \\<br /> yz = xyz/2 \Longrightarrow x = 2\\<br /> zx = xyz/3 \Longrightarrow y=3

That's Awesome ! Thanks a lot !