SOLUTION: How do you solve this system of three linear equations using elimination method 3x+2y+z=1 x+y+z=0 5x+3y-2z=-4

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Question 1196171: How do you solve this system of three linear equations using elimination method
3x+2y+z=1
x+y+z=0
5x+3y-2z=-4
Found 2 solutions by math_tutor2020, MathLover1:
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

I'll get you started.

There are a multiple number of ways to approach elimination problems.
It's mostly a trial and error type of thing, or something you develop a good eye for once you get enough practice.

Let's eliminate the variable z
The first two equations have +z in them
If we flip the signs of each term in equation (2), then we go from x+y+z = 0 to -x-y-z = 0

So we have
3x+2y+z = 1
-x-y-z = 0

Add the equations straight down
  • The x terms add to 3x + (-x) = 2x
  • The y terms add to 2y + (-y) = y
  • The z terms cancel because z + (-z) = 0z = 0, which we intended (and why we did the sign flip in the second equation)
  • The right hand sides add to 1+0 = 1
We end up with 2x+y = 1
I'll call this equation (4)

Return back to the original system
3x+2y+z=1
x+y+z=0
5x+3y-2z=-4


Temporarily delete the first equation to get
x+y+z=0
5x+3y-2z=-4

Now if we were to double everything in equation (2), then we go from x+y+z = 0 to 2x+2y+2z = 0

So this system
x+y+z=0
5x+3y-2z=-4

is the same as
2x+2y+2z=0
5x+3y-2z=-4

Add straight down:
  • The x terms add to 2x + 5x = 7x
  • The y terms add to 2y + 3y = 5y
  • The z terms cancel because 2z + (-2z) = 0z = 0
  • The right hand sides add to 0 + (-4) = -4
We end up with:
7x+5y = -4
which I'll refer to as equation (5)

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We now have a smaller system of equations
2x+y = 1
7x+5y = -4
which were equations (4) and (5) mentioned earlier.

I'll let you finish up the problem. Feel free to ask about any step, or if you are still stuck.

Hint: multiply equation (4) by -5 so you can eliminate the y variable.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

using elimination method
3x%2B2y%2Bz=1.....eq1
x%2By%2Bz=0......eq.2
5x%2B3y-2z=-4......,eq.3
start with
3x%2B2y%2Bz=1.....eq1
x%2By%2Bz=0......eq.2
__________________________subtract eq.2 from eq1
3x%2B2y%2Bz-x-y-z=1-0...........eliminate z

2x%2By=1......eq.a
now
x%2By%2Bz=0......eq.2
5x%2B3y-2z=-4......,eq.3
__________________________multiply eq1 by 2
2x%2B2y%2B2z=0......eq.2
5x%2B3y-2z=-4......,eq.3
__________________________add both
2x%2B2y%2B2z%2B5x%2B3y-2z=0-4..........eliminate 2z
7x%2B5y=-4..................eq.b
now
2x%2By=1......eq.a
7x%2B5y=-4..................eq.b
__________________________multiply eq.a by 5
10x%2B5y=5......eq.a
7x%2B5y=-4..................eq.b
__________________________subtract eq.b from eq.a
10x%2B5y-7x-5y=5-%28-4%29
3x=5%2B4
3x=9
x=3
go to
2x%2By=1......eq.a, plug in x=3
2%2A3%2By=1
6%2By=1
y=1-6
y=-5
now
x%2By%2Bz=0......eq.2, plug in x=3 and y=-5
3-5%2Bz=0
-2%2Bz=0
z=2
so, solution to this system is: x=3,y=-5, z=2