SOLUTION: x+y+z=-6
x-y+4z=6
5x+y+z=-26
Algebra.Com
Question 242750: x+y+z=-6
x-y+4z=6
5x+y+z=-26
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
x+y+z=-6
x-y+4z=6
5x+y+z=-26
-----------------
Subtract 1st from 2nd;
Subtract 5*1st from 3rd:
-------------------------------
x+y+z=-6
0x-2y+3z=12
0x-4y-4z=4
-----------------------
Simplify 3rd:
x+y+z=-6
0x-2y+3z=12
0x-2y-2z=2
--------------------
Subtract 2nd from 3rd:
x+y+z=-6
0x-2y+3z=12
0x+0y-5z=-10
---------------------
Solve 3rd to get z = 2
Substitute into 2nd to solve for "y": y = -3
Substitute into 1st to solve for "x": x = -5
=====================================================
Cheers,
Stan H.
RELATED QUESTIONS
Enter the augmented matrix.
7x+y+5z=2
5x-8y-z=6... (answered by richwmiller)
solve x + 3y + z = 1 2x – y – z = 6
5x + y + z = 1
using matrices. (answered by Fombitz)
1) x+y=9 x+2y=13 x= y=
2) 3x+7y= 24 9x+5y=24 x= y=
3) 4x+y=17 2x+3y+21 x= (answered by Fombitz)
X+y+z=66
X+3=y... (answered by ewatrrr)
Solve the system:
5x - 3y + z = -6
x + 3y - z = 0
4x + y + z =... (answered by checkley71)
How do you solve for x y z by elimination method
2x-3y+4z=6
5x+y+3z=14... (answered by josgarithmetic)
x+y=5
y+z=7... (answered by MathLover1)
Hi,-
I have a problem with substitution that does not allow the use of matrices:
2x + (answered by Alan3354)
-3x-y=-6
y-4z=11
x-2y+z=-7
solve the system of linear... (answered by josh's girl)