Question

how do you solve the system of equations 2x+y-3z=-3 x-2y+z=6 x-3y+3z=10

Answer 1

either with substitution , or elimination, or matrix methods ...

Answer 2

or, you could define the line of intersection of the first 2 planes and determine where it intersects the plane of the last one

Answer 3

I came up with the answers x=34 y=-14 z=0 they worked in the first equation but not in the other two equations.

Answer 4

then thats not a solution :)

Answer 5

I don't know if I am doing something wrong when I try to calculate the other two equations because it did work in one of them x-2y+z=6 works

Answer 6

when x=0 we have:

2y-6z=-6
-2y +z =6
-----------
5z = 0; z=0, y=-3

(0,-3,0)
----------------------

when y=0 we have

2x-3z=-3
-2x-2z=-12
------------
-5z = -15, z=3 , x=3

(3,0,3)
------------------------

the line between the points is therefore:

x = 3+3t
y = 0+3t
z = 3+3t
-----------------------

plug these into the last equation to determine the value of t

x-3y+3z=10

(3+3t)-3(3t)+3(3+3t)=10

3+3t-9t+9+9t=10

3t=-2 , t = -2/3
-------------------

therefore

x = 3+3(-2/3)
y = 0+3(-2/3)
z = 3+3(-2/3)

x = 3-2 = 1
y = 0-2 = -2
z = 3-2 = 1

with any luck this worked oout without my typos

Answer 7

2-2-3=-3
1+4+1=6
1+6+3=10

yep

Answer 8

thank you so much

Answer 9

i most likely used a method that you are not familiar with; but even so, the solution does not change :)