Solving a Linear equation with 3 unknown variables

[pietersandi_w]

3x + 4y + 2y = 1

The solutions for x, y, and z is { ( (1/3-4l-2m) | 3l | 2m ) }, where y = l and z = m.

I've tried this method, presupposing y = l and m = z, then it came to

I. l = (1 - 3x -2m) / 4

II. m = (1 - 3x -4l) / 2

If I try to put in either (I) or (II) to x, it would come to the first equation.

I don't have any idea what should I do next.

Thanks.

 

[Ray Vickson]

3x + 4y + 2y = 1

The solutions for x, y, and z is { ( (1/3-4l-2m) | 3l | 2m ) }, where y = l and z = m.

I've tried this method, presupposing y = l and m = z, then it came to

I. l = (1 - 3x -2m) / 4

II. m = (1 - 3x -4l) / 2

If I try to put in either (I) or (II) to x, it would come to the first equation.

I don't have any idea what should I do next.

Thanks.

You cannot do more, in general. The equation defines a 2-dimensional plane in three dimensions, so will have infinitely many point in it; that is, you have a region defined by two "free" variables. Assuming your equation was supposed to be 3x + 4y + 2z = 1 (you wrote 3x + 4y + 2y = 1 !)
we can just say that
$$z = \frac{1}{3}\, (1-4y - 2z)$$
I'm not sure whether or not you wrote this, because your use of parentheses was sloppy and so it is impossible to tell what gets divided by 3; and besides that you used some other weird notation whose meaning is unclear.

 

[NascentOxygen]

Ray, is your equation supposed to be
$$x = \frac{1}{3}\, (1-4y - 2z)$$

 

[Ray Vickson]

Ray, is your equation supposed to be
$$x = \frac{1}{3}\, (1-4y - 2z)$$

Yes. (He says, as he cringes and slaps forehead.)