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Sagot :
(1,3,-7)
1) Let's write the following system of linear equations, and then solve them by the elimination method.
The first thing to do is to proceed with the elimination of one variable. In this case, let's chose to start with
x-6y+2z=5 x -2
2x-3y+z=4
3x+4y-z=-2
2x-12y+4z=10
2x-3y+z=4
3x+4y-z=-2
Adding these two equations:
2x-3y+z=4
-2x+12y-4z=-10
----------------------------------
9y -3z =-6
Now getting the other equations, let's eliminate another variable, in this case, "x"
-2x+12y-4z=-10 x 3
3x+4y-z=-2 x 2
Adding them
-6x +36y -12z=-30
6x+8y-2z=-4
----------------------------
0 44y -14z=-34
2) Now that we have a simpler system of two variables "y" and " z" let's operate them. Choosing a factor that we can multiply and cancel out the y variable, in this case, let's pick the Least Common Multiple between 9 and 44 = 396, so let's multiply them by
9y -3z = -6 x 44
44y-14z=-34 x -9
396y -132z = -264
-396y+126z=-306
----------------------------
6z =-42
z=-7
Plugging into the equation 9y -3z =-6
9y-3(-7)=-6
9y +21 =-6
9y=27
y=3
And for x
x-6y+2z=5
x-6(3)+2(-7)=5
x+18-14=5
x+4=5
x=1
3) So the solution for that system is (1, 3, -7).
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