Im pretty sure you need to use substituion.
solve for z in terms of x and y then
put then in another equation and solve for x in terms of y then
put it in the last equation and solve for x
from there you can find y and z. the solution (x,y,z) should be the point at which the planes intersect...i think.
hope this helps
i tried and failed otherwise i would show you how
sorry about that im working on it now. I tried substition and combinations and here is the result (im off somewhere but i believe this should work)...
multiply by -1 for first equation and 2 for the second:
-1 1 -1 = -10
4 3 -4 = -42
10 8 5 = -10
13x + 12y = -41
same thing only different
-2 2 -2 = -20
-8 -12 8 = 84
10 8 5 = -10
-2y + 11z = 74
from those 2 equation we just found find what x and z equal both in terms of y and you get:
dang i thought it would work...sorry...this comes out to be y=-5.6 something. Good luck.
CAS means Computer Algebra System, Computer algebra system - Wikipedia, the free encyclopedia
So not pencil and paper.
Without CAS, you can use row-reduction. First do it with just a 2x2 matrix then do this one. I did the first row reductions below quick so need to check them:
Now: first row times -2, add to second and first row times -10 and add to third gives:
Second row times 1/5, add to first and second times 18/5 add to third gives:
Keep doing it until you get ones on the diagonal in the first three columns (divide the second row by 5 right).