$$(a+1)x+y+3z=1$$$$8x+2y+(a+3)z=2$$$$3x+y+2z=-1$$
This question can be calculated with Gauss-elimination and I want to take away the y by taking $-2$ from the middle and $-1$ from the top and keep the third one as it is, my question is how do I calculate it with a parameter?
$$ \left \{ \begin{array}{lcr} (a+1)x+y+3z & = & 1 \\ 8x+2y+(a+3)z & = & 2 \\ 3x+y+2z & = & -1 \end{array}\right . \\ \left \{ \begin{array}{lcr} 2(a+1)x+2y+6z & = & 2 \\ 8x+2y+(a+3)z & = & 2 \\ 3x+y+2z & = & -1 \end{array}\right . \\ \left \{ \begin{array}{lcr} 2(a-3)x+(3-a)z & = & 0 \\ 8x+2y+(a+3)z & = & 2 \\ 3x+y+2z & = & -1 \end{array}\right . \\ \left \{ \begin{array}{lcr} (a-3)(2x-z) & = & 0 \\ 8x+2y+(a+3)z & = & 2 \\ 3x+y+2z & = & -1 \end{array}\right . $$ So, from here, you can conclude that if $a = 3$, the system gets overdefined (number of variables exceeds number of equations), and since first equation will give you nothing, second and third equations can be restated taking into account $a = 3$ $$ \left \{ \begin{array}{lcr} 8x+2y+6z & = & 2 \\ 3x+y+2z & = & -1 \end{array}\right . $$ Subtract, doubled second from first $$ 2x+2z =4 $$ or $z = 2 - x$, substitute it to third equation back, so you'll get $$ y = -1 - 3x - 4+2x = -x-5 $$ As you may see, due to the overdefinition, you observed one possible scenario – parametric solution, i.e. when one of the variables can take any value, uniquely defining other two.
Now, let's assume that $a \ne 3$. In this case initial system can be restated as $$ \left \{ \begin{array}{lcr} 2x-z & = & 0 \\ 8x+2y+(a+3)z & = & 2 \\ 3x+y+2z & = & -1 \end{array}\right . $$ First equation yields $z = 2x$. Consider latter two $$ \left \{ \begin{array}{lcr} 8x+2y+2(a+3)x & = & 2 \\ 3x+y+4x & = & -1 \end{array}\right . \\ \left \{ \begin{array}{lcr} 2(a+7)x+2y & = & 2 \\ 7x+y & = & -1 \end{array}\right . \\ \left \{ \begin{array}{lcr} (a+7)x+y & = & 1 \\ 7x+y & = & -1 \end{array}\right . $$ Consider first equation minus second $$ ax+7x+y-7x-y = ax = 2 $$ Again, you can see that $a \ne 0$. If it is, first and second equations are inconsistent, so you won't get any solution from there.
And finally, you can find that $$ x= \frac 2a $$ Substitute it to the equations $$ y = -1 - 7x = -1 - \frac {14}a = -\frac {a+14}a $$ and recalling initial equation $z = 2x$ you can finally find that $$ \left [ \begin{array}{lc} a \notin \{0,3\} & \left \{ \begin{array}{ccr} x & = & \frac 2a \\ y & = & -\frac {a+14}a \\ z & = & \frac 4a \end{array} \right . \\ a = 3 & \left \{ \begin{array}{ccr} y & = & -x-5 \\ z & = & 2-x \end{array} \right . \end{array} \right . $$
Disclaimer
Even though this solution is overly filled with elementary things that could and even should have been easily omitted, I've solely decided that it might be useful for you. You can check answers provided here with Wolfram alpha by this link.