I'm studying Algebra and I'm now at topic 'System of equations with 3 variables'. I'm having a hard time with the following example: $$ \begin{cases} 2x + 2y + 3z = 10\\ 3x + y-z = 0\\ x + y + 2z = 6 \end{cases} $$ I've tried solving the system by using the first and second equation, but I got very different results from using the second and third first... Aren't those systems supposed to be resolved in any order, just "eliminating" variables until you get the results of the three unknowns? O maybe I just did the math in the wrong way?
Using the first way I got the result: $(x= 15, y= -33, z= -46/3)$
Looking at the worked solution in the book (beginning with second and the third equations) the results were: $(x= 0, y= 2, z= 2)$
Thanks a lot!
Cleverson
Well the answer should indeed be the same, since the same values should (in the end) satisfy the equations. It can happen that the equations have more than one solution, but this doesn't seem to have been a conclusion that either you or the book reached in this case.
So starting from the initial given equations, and basically eliminating $z$ then $y$, my solution would run something like:
$$ \begin{align} 2x + 2y + 3z &= 10 \tag{a}\\ 3x + y-z &= 0 \tag{b}\\ x + y + 2z &= 6 \tag{c}\\ \\ 11x+5y +0z&=10 \tag{d: a+3b}\\ 7x+3y+0z &=6 \tag{e: 2b+c}\\ \\ 2x+0y+0z &= 0 \tag{f: 5e-3d}\\ x &= 0 \tag{g} \\ \hline \\ 0x+3y+0z &=6 \tag{h: e-7g}\\ y &=2 \tag{i}\\ \hline \\ 0x+0y+2z &=4 \tag{j: c-i-g}\\ z &=2 \tag{k}\\ \hline \\ \end{align} $$
So: unique solutions, matching the book as expected. If you post your method of solution, someone might be able to identify where you went wrong.