# solve equation in 3 variables

• Dec 24th 2012, 11:45 PM
earthboy
solve equation in 3 variables

Solve for the set of equations :

$\displaystyle x+y+z=12 \\ xy+yz+zx=44 \\ xyz=48$

And...Merry Christmas & Happy new Year(Party)
• Dec 25th 2012, 01:33 AM
veileen
Re: solve equation in 3 variables
Merry Christmas! ^^

$\displaystyle xyz=48\Rightarrow x\neq 0, y\neq 0, z\neq 0$, knowing this we can multiply/ divide by x, y or z.

Starting from the first relation, we have:

$\displaystyle x+y+z=12\Rightarrow x(x+y+z)=12x\Rightarrow x^2+xy+xz=12x\Rightarrow$
$\displaystyle \Rightarrow x^2+xy+xz-(xy+yz+zx)=12x-44\Rightarrow x^2-yz=12x-44$

We know that $\displaystyle xyz=48$, so that, yz which is kind of a problem for us, can be written as $\displaystyle \frac{48}{x}$:

$\displaystyle x^2-\frac{48}{x}=12x-44 \Rightarrow x^3-48=12x^2-44x\Rightarrow x^3-12x^2+44x-48=0$

For x=2: $\displaystyle 2^3-12\cdot 2^2+44\cdot 2-48=8-48+88-48=0$. That means that we write our equation this way:

$\displaystyle x^3-12x^2+44x-48=x^3-2x^2-10x^2+20x+24x-48=$

$\displaystyle =x^2(x-2)-10x(x-2)+24(x-2)=(x-2)(x^2-10x+24)=0$

So we know that: $\displaystyle (x-2)(x-4)(x-6)=0$. x can then be 2, 4 or 6.
We can (and actually should) do the same thing for y and z, so y and z can also be 2, 4 or 6.

$\displaystyle (x, y, z) \in \left \{ (2, 4, 6); (2, 6, 4); (4, 2, 6); (4, 6, 2), (6, 2, 4); (6, 4, 2) \right \}$
• Dec 25th 2012, 06:08 AM
Soroban
Re: solve equation in 3 variables
Hello, earthboy!

Quote:

$\displaystyle \text{Solve: }\;\begin{Bmatrix}x+y+z\;=\;12 \\ xy+yz+zx\;=\;44 \\ xyz\;=\;48 \end{Bmatrix}$

$\displaystyle \text{We note that }x,y,z\text{ are the roots of the equation:}$
. . . $\displaystyle u^3 - 12u^2 + 44u - 48 \:=\:0$

$\displaystyle \text{Factor: }\:(u-2)(u-4)(u-6) \:=\:0$

$\displaystyle \text{Therefore: }\:\{x,y,z\} \:=\:\{2,4,6\}\text{ in some order.}$