Question

Solve the following system of equations.

2x + 3y - z = 1
3x + y + 2z = 12
x + 2y - 3z = -5

Answer 1

ick i would cheat

Answer 2

otherwise it is a real pain to use elimination repeatedly
do you have to do it step by step?

Answer 3

http://www.wolframalpha.com/input/?i=2x+%2B+3y+-+z+%3D+1%2C+3x+%2B+y+%2B+2z+%3D+12%2C+x+%2B+2y+-+3z+%3D+-5

Answer 4

2x+3y-z=0 ...(1)
3x+y+2z=12 ...(2)
x+2y-3z=-5 ...(3)
from (1) 2x+3y=z,substitute in (2) and (3)
3x+y+2(2x+3y)=12
or 7x+7y=12 ...(4)
x+2y-3(2x+3y)=-5
-5x-7y=-5 ...(5)
adding (4) and (5)
2x=7
x=7/2
from (4)
7*7/2+7y=12
49+14y=24
14y=24-49=-25
y=-25/14
substitute the values of x and y and get the value of z

Answer 5

2x + 3y - z = 1
3x + y + 2z = 12 -->(-3)
----------------
2x + 3y - z = 1
-9x -3y - 6z = -36 (result of multiplying by -3)
---------------
-7x - 7z = -35

3x + y + 2z = 12 -->(-2)
x + 2y - 3z = -5
---------------
-6x - 2y - 4z = -24 (result of multiplying by -2)
x + 2y - 3z = -5
---------------add
-5x - 7z = - 29

-7x - 7z = -35 -->(-1)
-5x - 7z = -29
----------------
7x + 7z = 35 (result of multiplying by -1)
-5x - 7z = -29
----------------add
2x = 6
x = 3

7x + 7z = 35
7(3) + 7z = 35
21 + 7z = 35
7z = 35 - 21
7z = 14
z = 2

2x + 3y - z = 1
2(3) + 3y - 2 = 1
6 - 2 + 3y = 1
4 + 3y = 1
3y = 1 - 4
3y = -3
y = -1

check...
3x + y + 2z = 12
3(3) + (-1) + 2(2) = 12
9 - 1 + 4 = 12
8 + 4 = 12
12 = 12 (correct)

SOLUTION : x = 3, y = -1, z = 2

Answer 6

@Brianna09201995 ...sorry it took so long

Answer 7

I kept messing up, but I finally got it :)

Answer 8

any questions ?

Answer 9

sorry i wrote the wrong statement