SOLUTION: How do you solve the system of three linear equations using the elimination method which theses equations 2x-4y+5z=-33 4x-y=-5 -2x+2y-3z=19

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Question 1196110: How do you solve the system of three linear equations using the elimination method which theses equations
2x-4y+5z=-33
4x-y=-5
-2x+2y-3z=19
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
system%282x-4y%2B5z=-33%2C%0D%0A4x-y=-5%2C%0D%0A-2x%2B2y-3z=19%29

Since z is already eliminated from the 2nd equation,
let's eliminate z from the 1st and 3rd equations:

Multiply the first equation by 3 and the 3rd equation by 5,
so the coefficients of z will be alike with opposite signs:



Add the equations term by term and the 15z and the -15z will cancel:

-4x-2y=-4

Put that with the original 2nd equation:

system%28-4x-2y=-4%2Cmatrix%281%2C3%2C%22%22%2C%22%22%2C4x-y=-5%29%29+

Add those term by term, the 4x and the -4x will cancel:

-3y=-9%0D%0A%0D%0A%7B%7B%7By=3

Substitute 3 for y in 

4x-y=-5

4x-3=-5

4x+=+-2

x=-1%2F2

Substitute -1/2 for x and 3 for y in 2x-4y+5z=-33

2x-4y%2B5z=-33
2%28-1%2F2%29-4%283%29%2B5z=-33
-1-12%2B5z=-33
-13%2B5z=-33
5z=-20
z=-4

(x,y,z) = (-1/2,3,-4}}}

Edwin