I want to solve the following euqations:
$6y-2xz = 0 $
$6x-2yz = 0 $
$x^2 + y^2 -8 = 0 $
One way would be to multiply the first equations with y and the second with x and then substract the first from the socond - but what, if I want to solve more complicated equations? If i try to solve y out of the first equation and substitute it in the second i get:
$x(6 - 2z/6)=0$
I cant substitute x in the third equation.
Thanks for your help!
You can generally solve a system of equations by solving the first equation for one of the variables, and substituting for it into the remaining equations. This will turn a system of $m$ equations in $n$ unknowns into a system of $m - 1$ equations in $n -1$ unknowns. To which you can again do the same thing, until finally you run out of equations or unknowns. If $n = m$, then you get one equation in one variable, which you can then solve, and back substitute the values backwards.
For your system, solving the first equation for $y$ gives: $$y = \frac{xz}3$$ (I recommend simplying as far as possible before you substitute into the other equations - it will make your life easier). Substituting for $y$ into the remaining equations gives: $$6x - 2(\frac{xz}3)z = 2x\left(3 - \frac{z^2}3\right) = 0$$ $$x^2 + (\frac{xz}3)^2 - 8 =x^2\left(1 + \frac{z^2}9\right) - 8 = 0$$
With some minor simplifications
$$x\left(9 - z^2\right) = 0$$ $$x^2(9 + z^2) = 72$$ The top equation has a product of two factors equal to $0$. Recall that the only way this can be is if one of the factors is $0$, so the first equation requires that either $x = 0$ or $9 - z^2 = 0$. Now substitute $x = 0$ into the bottom equation gives $0(9 + z^2) = 72$, which cannot be. So $x = 0$ is not a solution. But if $z^2 = 9$, you can substitute for $z$ into the bottom equation to get $x^2(18) = 72$, which can be solved for $x$. Finally, substitute your solutions for $x$ and $z$ back into $y = \frac{xz}3$ to find the value of $y$.
You should find $4$ separate solutions $(x, y, z)$ of the original system of equations.