The idea is to start with a system of 3 equations in 3 variables, then
eliminate a variable and get a new system of only 2 equations in only 2
variables. Then we eliminate one of those and we end up with only 1 equation
and 1 variable, (which is the first kind of equation we ever learned to solve
when we just began studying algebra.)
9x + 5y - z = -11
6x + 4y + 2z = 2
2x - 2y + 4z = 4
Notice that it just so happens that the 2nd and 3rd equations can be divided
through by 2, so we might as well do that from the start:
9x + 5y - z = -11
3x + 2y + z = 1
x - y + 2z = 2
Pick a letter to eliminate and pick two of the equations that contain that
letter to eliminate it from. It is arbitrary which you pick, but you should
pick them so as to minimize the amount of work. I will pick the first two
equations to eliminate z from:
We don't need to multiply either by anything because the terms in z will cancel
if we add them -- corresponding term to corresponding term.
9x + 5y - z = -11
3x + 2y + z = 1
------------------
12x + 7y = -10
Now we need another equation without a z-term. So we pick the original
equation that we haven't used and one of the ones we already used. We have not
used this equation, (the third original equation):
x - y + 2z = 2
We will take it and the second original equation
x - y + 2z = 2
3x + 2y + z = 1
(At this point BE SURE to eliminate the SAME variable z that was eliminated the
first time). We must eliminate z from these too, so we will multiply the
bottom one by -2 to make the z-terms cancel when we add them term by term.
x - y + 2z = 2
-2(3x + 2y + z = 1)
x - y + 2z = 2
-6x - 4y - 2z = -2
-----------------
-5x - 5y = 0
This can be divided through by -5
x + y = 0
Now we have a system of two equation and two variables:
x + y = 0
12x + 7y = -10
We can solve them any way we know how, but since you said elimination I will
not switch over to substitution just yet. Eliminate the y term by multiplying
the first one through by -7 before adding term by term.
-7(x + y = 0)
12x + 7y = -10
-7x - 7y = 0
12x + 7y = -10
--------------
5x = -10
x = -2
Now we switch to substitution. Substitute -2 for x in
x + y = 0
-2 + y = 0
y = 2
Now substitute -2 for x and 2 for y in one of the original equations, say the
second original equation:
3x + 2y + z = 1
3(-2) + 2(2) + z = 1
-6 + 4 + z = 1
-2 + z = 1
z = 3
Solution: (x,y,z) = (-2,2,3)
Edwin
