So I was again trying to come up with questions that I thought that I might be able to solve when I came up with this problem: $$\text{Find the points }(x,y,z)\text{ that will satisfy the equation }6x-3y+9z=18$$which I thought that I might be able to do. Here is my attempt for solving the $3$-variable equation:$\color{white}{\require{\cancel}{.}}$ $$6x-3y+9z=18$$$$-3y=18-6x-9z$$$$y=-6+2x+3z$$$$6x=18+3y-9z$$$$x=3+0.5y-1.5z$$$$9z=18-6x+3y$$$$z=2-\dfrac{2}{3}x+\dfrac{1}{3}y$$$$\cancel{18}+3y-9z\cancel{-18}-6x-9z+18-6x+3y=18$$$$-12x+6y-18z=0$$$$6y=12x+18z$$$$y=2x+3z$$$$-12x=-6y+18z$$$$12x=6y-18z$$$$x=0.5y-1.5z$$$$-18z=-6y+12x$$$$18z=6y-12x$$$$z=\dfrac{1}{3}y-\dfrac{2}{3}x$$$$0.5y-1.5z=0$$$$y-3z=0$$$$y=3z$$$$x-y-z=0.5y-1.5z-2x-3z-\dfrac{1}{3}y+\dfrac{2}{3}x$$$$-8x+y-27z=0$$$$y=8x+27z+18$$$$-8x=18+27z-y$$$$x=-2.25-3.375z+0.125y$$$$18+\cancel{27z}\cancel{-27z}+y+8x+27z+\cancel{18}=\cancel{18}$$$$y=-18-8x-27z$$However, I realized that the equation had become an infinite loop of nothingness and decided to plug it into Wolfram Alpha, which gave me$$y=3n+2x$$$$z=n+2$$And plugging this into Desmos and playing around with the sliders, we get:$$x=-8$$$$y=14$$$$z=12$$Meaning that the $(x,y,z)$ coordinate solution is$$(-8,14,12)$$
My question
Is the solution that I have achieved correct, or what could I do to attain it more easily?
To clarify
I have done another attempt on this question before this one, however, even though my solution was correct in this case $(2.25,-1.5,0)$, I redid the entire equation since I found out that I had accidently removed a variable when there was nothing cancelling it out in the first place.
You have a single linear equation with three variables, so there are infinitely many solutions. The solution points $(x,y,z)$ constitute a 2D plane in 3D space. These observations are the beginnings of linear algebra. Also see systems of linear equations.