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.

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- Jul 19th 2006, 08:08 PMkwtolleySystem 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, 08:28 PMCaptainBlackQuote:

Originally Posted by**kwtolley**

you have is true then the candidate is the solution, so:

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

The LHS of the second equation is $\displaystyle (2\times 1)-1+(3 \times 6)=2-1+18=19$,

which is not equal to the RHS, so no $\displaystyle (1,1,6)$ is not a solution

of the system.

RonL - Jul 20th 2006, 12:50 AMkwtolleyrework
I'm going to re work the problem and repost, is that ok. Thanks for checking my answer.