Is there a particular method they want you to use?
Here I'll show elimination.
Probably the easiest variable to eliminate is x.
Multiply the 2nd equation by -3. When adding
this result to the 1st equation, the x terms drop out.
-3*(x+2y-3z) = -3*-6
-3x-6y+9z = 18
Now, add this to equation 1:
-3x-6y+9z = 18
+ 3x-4y+2z = -5
--------------------------
-10y+11z = 13 (no x term now)
Next, we want to multiply original eqn. 2
by 2. We do this because when added to
eqn. 3, the x terms will drop out.
2*(x+2y-3z) = 2*-6
2x+4y-6z = -12
Now, add to eqn. 3
2x+4y-6z = -12
+ -2x-3y+5z = 11
--------------------------
y-z = -1 (no x term now)
So, now we have two equations with
only two variables (y & z)
-10y+11z = 13
y - z = -1
Now, we need to eliminate either y or
z. I'll choose to eliminate y. We can
do this by multiplying the bottom eqn by
10 and adding to the top.
10*(y-z) = 10*(-1)
10y-10z = -10
Now, adding the two..
10y-10z = -10
+ -10y+11z = 13
-------------------------
z = 3 (y drops out)
Now, we can plug in z into one of
the previous equations to find y. I'm
going to choose y-z=-1 as that is easiest.
y-z=-1
y-(3)=-1
y-3=-1
y-3=-1
+3 +3 add 3 to both sides
-------------
y = 2
Now all we need is x. Can substitute y and
z into one of the previous equations. I'll
choose the 2nd equation from the beginning.
x+2y-3z = -6
x + 2(2) - 3(3) = -6
x + 4 - 9 = -6
x - 5 = -6
+ 5 +5 add 5 to both sides
-----------------
x = -1
So, our solution is (-1, 2,3)
Plug these values into the original
equations to make sure they check
