0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 '15 at 14:49
  • $\begingroup$ substituting your values no one of the equations is satisfied, so redo your work.. $\endgroup$ – Emilio Novati Mar 25 '15 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. $\endgroup$ – Rory Daulton Mar 25 '15 at 18:17
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.