Thread: Solving for a system of equations

1. Solving for a system of equations

The problem is as follows:

Sometimes we may need to solve a system of equations that isn't linear, but wants to be. With this in mind, solve the following system of nonlinear equations for x, y, and z.

xz + yz + xy = -1
xz - yz + 2xy = 5
xz - 2yz + 2xy = 7

I put the equations into matrices and got so far as xz = 20, -yz = -5, -xy = -16. But now I don't know where to go... any help would be much appreciated!

2. Re: Solving for a system of equations

Hey widenerl194.

Hint: Try getting y = f(x,z), z = g(x,y) and then substitute both to get a function of h(x) = 0. (Example x = 20/z).

3. Re: Solving for a system of equations

Note that if you use the first equation to get xz = -1 - yz - xy

plug it into eqn 2 and 3.

to get (2) -1 - yz - xy - yx + 2xy = 5 or (2) -2yz + xy = 6
and again for (3) -1 -yz - xy - 2yz + 2xy = 7 or (3) -3yz + xy = 8
use yz=p and xy=q to solve the linear system
$\begin{bmatrix}-2 & 1 \\ -3 & 1 \end{bmatrix} x = \begin{bmatrix} 6 \\ 8 \end{bmatrix}$
solving it you get p = -2, q = 2, or yz = -2 and xy = 2.
Plugging this information into any 3 of your original equations gives you xz = -1

so you got to solve 3 equations
(1)xz = -1
(2)yz=-2
(3)xy = 2

taking the first one $z = \frac{-1}{x}$ plug into (2) to get $y \frac{-1}{x} = -2$
taking 3, $x = \frac{2}{y}$
so $y \frac{1}{\frac{2}{y}} = 2$ or the solution to $y^2 = 4$ which is $y = +2, -2$

now working backwords we get x = +1 or -1, z = +1 or -1.
Now since xz = -1, set x = -1, z = 1 or x = 1 and z = -1
and since yz = -2, since z = 1, y = -2 or z = -1 and z = 2

so the solution set is (x,y,z) where x=+1 or -1, y = +2 or -2 and z = +1 or -1 with the only restriction that the sign of x,z are reverse (e.g if x = 1, then z = -1) and the sign of y and z are reverse (e.g if z = -1, then y = 2)

The solution set is (x,y,z) = $\{(-1,-2,1), (1,2,-1) \}$