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(3817) (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)
------------------------------------------------------
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) (Show Source): You can put this solution on YOUR website!
using elimination method
.....eq1
......eq.2
......,eq.3
start with
.....eq1
......eq.2
__________________________subtract eq.2 from eq1
...........eliminate
......eq.a
now
......eq.2
......,eq.3
__________________________multiply eq1 by
......eq.2
......,eq.3
__________________________add both
..........eliminate
..................eq.b
now
......eq.a
..................eq.b
__________________________multiply eq.a by
......eq.a
..................eq.b
__________________________subtract eq.b from eq.a
go to
......eq.a, plug in
now
......eq.2, plug in and
so, solution to this system is: ,,
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