Question

Use the elimination method to solve the following system of equations.

2x - y + z = -3
2x + 2y + 3z = 2
3x - 3y - z = -4

A. (-1, -3, -2)
B. (1, -3, -2)
C. (1, 3, -2)
D. (1, 3, 2)

Answer 1

@Yttrium

Answer 2

@texaschic101 Can you help me with this?

Answer 3

2x - y + z = -3
2x + 2y + 3z = 2 --->(-1)
-------------
2x - y + z = -3
-2x - 2y - 3z = -2 (results of multiplying by -1)
-------------add
-3y - 2z = -5

2x + 2y + 3z = 2 -->(-3)
3x - 3y - z = -4 -->(2)
-------------------
-6x - 6y - 9z = -6 (results of multiplying by -3)
6x - 6y - 2z = -8 (results of multiplying by 2)
------------------add
-12y - 11z = - 14

-3y - 2z = -5 --->(-4)
-12y - 11z = -14
-----------------
12y + 8z = 20 (results of multiplying by -4)
-12y - 11z = -14
----------------add
-3z = 6
z = -2

-3y - 2z = -5
-3y - 2(-2) = -5
-3y + 4 = -5
-3y = -5 - 4
-3y = -9
y = 3

2x - y + z = -3
2x - 3 -2 = -3
2x - 5 = -3
2x = 5 - 3
2x = 2
x = 1

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

x = 1, y = 3 and z = -2 (C)

Answer 4

not that hard...just VERY time consuming

Answer 5

Thank You I'm having trouble with the systems of 3 equations. @texaschic101

Answer 6

I can see why. If you make any mistakes at the beginning, it will carry through to the end and mess up your whole problem. You have to make sure you get the signs correct. It can be a pain in the butt sometimes...lol