Question

Solve each system by elimination.

x + y - 2z = 8
5x - 3y + z = -6
-2x - y + 4z = -13

Answer 1

dealing with the 1st and 3rd equation....I picked these two because when adding them, the y's will cancel out...
x + y - 2z = 8
-2x - y + 4z = -13
----------------add
-x + 2z = - 5

now take the 2nd equation and either of the other two equations and eliminate the y's out. It has to be the same variable you eliminated in the first equations you used.
5x - 3y + z = -6
x + y - 2z = 8 -->(3)
-----------------
5x - 3y + z = -6
3x + 3y - 6z = 24 (result of multiplying by 3)
-----------------add
8x - 5z = 18

Now take the equations you got after you eliminated the y's and eliminate another variable.
-x + 2z = -5 -->(8)
8x - 5z = 18
---------------
-8x + 16z = -40 (result of multiplying by 8)
8x - 5z = 18
--------------add
11z = - 22 -- divide by 11
z = -2

now sub -2 in for z in either equation that is missing the y's
-x + 2z = -5
-x + 2(-2) = -5
-x - 4 = -5
-x = -5 + 4
-x = -1 -- multiply by -1 to make x positive
x = 1

now sub -2 in for z and 1 in for x in any of the original equations
x + y - 2z = 8
1 + y - 2(-2) = 8
1 + y + 4 = 8
y + 5 = 8
y = 8 - 5
y = 3

check...
5x - 3y + z = -6
5(1) - 3(3) + (-2) = -6
5 - 9 - 2 = -6
5 - 11 = -6
-6 = -6 (correct)

x = 1, y = 3 and z = -2

any questions ?

Answer 2

No, this makes perfect sense. Thank you so much!