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 whatOriginally 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