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 !