# Solving a system of equations .

1. Aug 8, 2013

1. The problem statement, all variables and given/known data

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

2. Relevant equations
-

3. 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.

2. Aug 8, 2013

### 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.

3. Aug 8, 2013

### Ray Vickson

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.

4. Aug 9, 2013

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 ?

5. Aug 9, 2013

### Ray Vickson

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
$$6b+6c=5p\\ 21a+21c=10p \\ 14a+14b=9p$$
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
$$xy = xyz/7 \Longrightarrow z = 7 \\ yz = xyz/2 \Longrightarrow x = 2\\ zx = xyz/3 \Longrightarrow y=3$$

6. Aug 10, 2013