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!


  • $\begingroup$ Both ways should give you the same result, I suppose there may be some mistakes in the working that lead you to a different answer. $\endgroup$
    – LaBird
    Mar 25, 2015 at 14:49
  • $\begingroup$ substituting your values no one of the equations is satisfied, so redo your work.. $\endgroup$ Mar 25, 2015 at 15:13
  • $\begingroup$ It is very hard for us to help you without more information, especially the details of your work. There are many ways to solve systems of linear equations, and there are even many variations within the "elimination of variables" method you mention. $\endgroup$ Mar 25, 2015 at 18:17

1 Answer 1


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.


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