4x−y+2z=11
x+2y−z=−1
2x+2y−3z=−1
can someone help me find x,y,z, I have tried isolation but i cant seem to get it
2 answers
never mind guys i got it cuz im smart and i never give up.
I would use elimination, especially after looking at the y terms
#3 - #2: x - 2z= 0 , so we already know z = 2y
#1 times 2 : 8x - 2y + 4z = 22
original #2 as is: x + 2y - z = -1
add them:
9x + 3z = 21
or
3x + z = 7
double this: 6x + 2z = 14
using with : x - 2z = 0
add them
7x = 14
x = 2
in 3x+z=7 ----> 6+z=7
z = 1
in the original #2:
x+2y-z=-1
2+2y-1=-1
2y=-2
y=-1
x=2 , y=-1, z=1
The above solution is not unique.
Other than choosing the y's, there was really no plan, you just sort of go with the flow.
#3 - #2: x - 2z= 0 , so we already know z = 2y
#1 times 2 : 8x - 2y + 4z = 22
original #2 as is: x + 2y - z = -1
add them:
9x + 3z = 21
or
3x + z = 7
double this: 6x + 2z = 14
using with : x - 2z = 0
add them
7x = 14
x = 2
in 3x+z=7 ----> 6+z=7
z = 1
in the original #2:
x+2y-z=-1
2+2y-1=-1
2y=-2
y=-1
x=2 , y=-1, z=1
The above solution is not unique.
Other than choosing the y's, there was really no plan, you just sort of go with the flow.