# System of equation

• Jul 19th 2006, 09:08 PM
kwtolley
System of equation
The solution to the system of equations below is
x+2y-z=-3
2x-y+3z=14
x+4y-2z=-8
My answer after using excel is (1,1,6) is this right thanks for checking.
• Jul 19th 2006, 09:28 PM
CaptainBlack
Quote:

Originally Posted by kwtolley
The solution to the system of equations below is
x+2y-z=-3
2x-y+3z=14
x+4y-2z=-8
My answer after using excel is (1,1,6) is this right thanks for checking.

Substitute the candidate solution back into the equations, then if what
you have is true then the candidate is the solution, so:

The LHS of the first equation is: $1+(2\times 1)-6=-3$ checks.

The LHS of the second equation is $(2\times 1)-1+(3 \times 6)=2-1+18=19$,
which is not equal to the RHS, so no $(1,1,6)$ is not a solution
of the system.

RonL
• Jul 20th 2006, 01:50 AM
kwtolley
rework
I'm going to re work the problem and repost, is that ok. Thanks for checking my answer.