Find conditions on $a$, $b$, $c$, and $d$ for which the following system has solutions:
$$2x+4y+z+3w=a $$ $$-3x+y+2z-2w=b $$ $$12x+5y-4z+12w=c $$ $$13x+10y-z+13w=d$$
I got the system down to:
Step 1: $$Eq4↔︎Eq1$$ $$Eq1→Eq1+2Eq2$$ $$Eq2→Eq2+3Eq3$$ $$Eq3→Eq3+4Eq2$$ $$13x+10y-z+13w=d$$ $$12y+5z+7w=b+3a$$ $$9y+4z+4w=c+4b$$ $$6y+5z-w=a+2b$$ Step 2: $$Eq4↔︎Eq2$$ $$Eq2→3Eq2-4Eq3$$ $$Eq3→4Eq3-3Eq2$$ $$13x+10y-z+13w=d$$ $$6y+5z-w+a+2b$$ $$z-5w=-9a+13b+4c$$ $$-z+5w=9a-13b-4c$$ Step 3: $$Eq4→Eq4+Eq3$$ $$13x+10y-z+13w = d $$ $$6y+5z-w = a+2b $$ $$z-5w =-9a+13b+4c $$ $$0=0$$
Would my answer be equation $3$?
There was an example in class where the last equation was (for example) $0=a+b+c$, and that was the answer.
As for where you went wrong: I think you tried to apply multiple row operations simultaneously (which results in an inequivalent system of equations). However, row operations should be applied one at a time, and order is important. Additionally, the choice of row operation should lead us closer to row echelon form.
Let's look at the proposed row operations for "step 1" performing one after the other:
\begin{align*} \begin{bmatrix} 2 & 4 & 1 & 3 & a \\ -3 & 1 & 2 & -2 & b \\ 12 & 5 & -4 & 12 & c \\ 13 & 10 & -1 & 13 & d \\ \end{bmatrix} & \xrightarrow{R_1 \leftrightarrow R_4} \begin{bmatrix} 13 & 10 & -1 & 13 & d \\ -3 & 1 & 2 & -2 & b \\ 12 & 5 & -4 & 12 & c \\ 2 & 4 & 1 & 3 & a \\ \end{bmatrix} \\ & \xrightarrow{R_1 \gets R_1 + 2R_2} \begin{bmatrix} 7 & 12 & 3 & 9 & 2b+d \\ -3 & 1 & 2 & -2 & b \\ 12 & 5 & -4 & 12 & c \\ 2 & 4 & 1 & 3 & a \\ \end{bmatrix} \\ & \xrightarrow{R_2 \gets R_2 + 3R_3} \begin{bmatrix} 7 & 12 & 3 & 9 & 2b+d \\ 33 & 16 & -10 & 34 & b+3c \\ 12 & 5 & -4 & 12 & c \\ 2 & 4 & 1 & 3 & a \\ \end{bmatrix} \\ & \xrightarrow{R_3 \gets R_3 + 4R_2} \begin{bmatrix} 7 & 12 & 3 & 9 & 2b+d \\ 33 & 16 & -10 & 34 & b+3c \\ 144 & 69 & -44 & 148 & 4b+13c \\ 2 & 4 & 1 & 3 & a \\ \end{bmatrix} \end{align*}
It looks nothing like what you seem to be aiming for (i.e., 0s in the right spots).
Here's what I get using Gaussian elimination: \begin{align*} \begin{bmatrix} 2 & 4 & 1 & 3 & a \\ -3 & 1 & 2 & -2 & b \\ 12 & 5 & -4 & 12 & c \\ 13 & 10 & -1 & 13 & d \\ \end{bmatrix} & \xrightarrow{R_2 \gets R_2+\tfrac{3}{2} R_1} \begin{bmatrix} 2 & 4 & 1 & 3 & a \\ 0 & 7 & \tfrac{7}{2} & \tfrac{5}{2} & \tfrac{3}{2}a+b \\ 12 & 5 & -4 & 12 & c \\ 13 & 10 & -1 & 13 & d \\ \end{bmatrix} \\ & \xrightarrow{R_3 \gets R_3-6 R_1} \begin{bmatrix} 2 & 4 & 1 & 3 & a \\ 0 & 7 & \tfrac{7}{2} & \tfrac{5}{2} & \tfrac{3}{2}a+b \\ 0 & -19 & -10 & -6 & -6a+c \\ 13 & 10 & -1 & 13 & d \\ \end{bmatrix} \\ & \xrightarrow{R_4 \gets R_4-\tfrac{13}{2} R_1} \begin{bmatrix} 2 & 4 & 1 & 3 & a \\ 0 & 7 & \tfrac{7}{2} & \tfrac{5}{2} & \tfrac{3}{2}a+b \\ 0 & -19 & -10 & -6 & -6a+c \\ 0 & -16 & -\tfrac{15}{2} & -\tfrac{13}{2} & -\tfrac{13}{2}a+d \\ \end{bmatrix} \\ & \xrightarrow{R_3 \gets R_3+\tfrac{19}{7} R_2} \begin{bmatrix} 2 & 4 & 1 & 3 & a \\ 0 & 7 & \tfrac{7}{2} & \tfrac{5}{2} & \tfrac{3}{2}a+b \\ 0 & 0 & -\tfrac{1}{2} & \tfrac{11}{14} & -\tfrac{27}{14}a+\tfrac{19}{7}b+c \\ 0 & -16 & -\tfrac{15}{2} & -\tfrac{13}{2} & -\tfrac{13}{2}a+d \\ \end{bmatrix} \\ & \xrightarrow{R_4 \gets R_4+\tfrac{16}{7} R_2} \begin{bmatrix} 2 & 4 & 1 & 3 & a \\ 0 & 7 & \tfrac{7}{2} & \tfrac{5}{2} & \tfrac{3}{2}a+b \\ 0 & 0 & -\tfrac{1}{2} & \tfrac{11}{14} & -\tfrac{27}{14}a+\tfrac{19}{7}b+c \\ 0 & 0 & \tfrac{1}{2} & -\tfrac{11}{14} & -\tfrac{43}{14}a+\tfrac{16}{7}b+d \\ \end{bmatrix} \\ & \xrightarrow{R_4 \gets R_4+R_3} \begin{bmatrix} 2 & 4 & 1 & 3 & a \\ 0 & 7 & \tfrac{7}{2} & \tfrac{5}{2} & \tfrac{3}{2}a+b \\ 0 & 0 & -\tfrac{1}{2} & \tfrac{11}{14} & -\tfrac{27}{14}a+\tfrac{19}{7}b+c \\ 0 & 0 & 0 & 0 & -5a+5b+c+d \\ \end{bmatrix} \end{align*} This is in row echelon form. Thus, we conclude that this system of equations is consistent if and only if $-5a+5b+c+d=0$ (agreeing with Rene Schipperus's answer). In fact, we can further say that there are infinitely many solutions whenever $-5a+5b+c+d=0$.