[SOLVED] 3 variables equation

yes this is homework , and NO im not looking for a solution , just for some points.

im trying to solve this for the last couple of hours , to no avail. any idea how to do the following (NOT SOLUTION!!!!!) : equation : Ax + By + Cz = D read A , B , C , D , and N , and find all the solutions in the range of : -N<= x,y,z <=N

for example : if x = k , y = f , z = m , so if -N<= k,f,m <=N so one of the solutions is : (k,f,m).

thanks!

Usually you need n equations to solve a system for n variables... So, I guess you have 3 equations or otherwise, you'll have to use parametrical values.

Have you looked at Gaussian elimination?

3 nested for loops?

What do you mean by

StOoZ said:

-N<= x,y,z <=N

pp. said: What do you mean by

Pretty sure that means find all solutions where x, y, and z are between +- N

cszikszoy said: Pretty sure that means find all solutions where x, y, and z are between +- N

Ah. That would be

(-N <= x <= N) v (-N <= y <= N) v (-N <= z <= N)

I guess.

For the problem to make any kind of sense, I also would guess that for x, y and z we have to consider natural numbers only?

For smallish values of N I would be tempted to do a brute force approach. Simply create all combinations of x,y,z and run each triplet through the equation.

in this situation (since we know the range) brute force is the only option.

ok the problem was solved , thanks a lot!! I just tried over and over again. ):P

You have 4 variables, a,b,c,d:

ax +by +cz = d

this is ax + by + cz -d = 0 which is a plane. This means the solution is something from analytic geometry (vectors).

You either need 3 equations (if d is given), or then you parametrize with t, where the number of variables minus the number of equations equals the degrees of free"doom".

Now assuming you have: 1x + 1y + 1z = 2 4x + 2y + 1z = 7 1x - 1y + 1z = -2

so you get the coefficient's matrix: x y z | d ------------------- |--------- 1 1 1 | 2 4 2 1 | 7 1 -1 1 | -2

-----------------------------

yo shall now substitute by row 2 = row 2 - (4/1) * row 1 and row 3 = row 3 - (1/1) * row 1

then row 3 = row 3 - (-1/[row2 y from last step]) * row 2

this leavs you with a triangular matrix: x y z | d ------------------- |--------- 1 1 1 | 2 0 2 3 | 1 0 0 3 | -3

-----------------------------

now you can do backward substitution: 3z = -3 --> z use the value obtained for z from row3 in row 2 --> y now use z & y from the two previous rows to solve for z in the first row.

Doing this in a computer program shouldn't be that difficult, just remember to use double and not int, because you're doing divisions... ):P

if you have one or more rows of zeros, or one line being a multiple of the other, then the system is unsolvable...

• exactly one solution for rank (koefficient matrix)= rank(extended coefficient matrix)
• infinite solutions for rank(coefficient matrix ) = rank(extended coefficient matrix)
• no solution if rank(coefficient matrix) < rank(extended coefficient matrix)

Now copy paste the rank function from somewhere from the internet... (and look at how it works)

also x[i] = det(A[i])/det(A) wikipedia, cramer's rule

now assume a parametrization: 4x + 2y + (a+2) z = a 2x + 2y + (a+1) z = 2a 5x + 4y - (a^2-a-2)z = 3a -1

now set up the extended coefficient matrix:

x y z | d ---------------------|--------- 4 2a a+2 | a 2 2a a+1 | 2a 4 4a -a^2+a+2 | 3a-1

-----------------------------

now subtract the first row (2/4) times from the second row and so forth

now this leaves: x y z | d ---------------------|--------- 4 2a a+2 | a 0 2a a | 2a 0 0 a*(a+1) | a-1

-----------------------------

so for a = -1 follows, rank(A) = rank(B) = 2 so you have infinite solutions

for a = 0 you have rank(A) = 1 and rank(B) = 2

so there exist no solutions for a = 0.

in every other case, you've got exactly one solution. Done. now check whether all solutions are in the range! implement! :lolflag: