$$\begin{split}\simplify[!noLeadingminus,unitFactor]{{a}x+{b}y} &\,=\var{c} \qquad\qquad&(1)\\ \simplify[!noLeadingminus,unitFactor]{{a1}x +{b1}y}  &\,=\var{c1} \qquad\qquad&(2)\end{split}$$

To get a solution for $x$, if we multiply equation (1) by $\var{abs(b1)}$ and equation (2) by $\var{abs(b)}$, we will have two equations with equal and opposite $y$-coefficients:

$$\begin{split} \simplify[!noLeadingminus,unitFactor]{{a*abs(b1)}x+{b*abs(b1)}y} &\,=\var{c*abs(b1)} \qquad\qquad&(3)\\\simplify[!noLeadingminus,unitFactor]{{a1*abs(b)}x +{b1*abs(b)}y}  &\,=\var{c1*abs(b)} \qquad\qquad&(4) \end{split}$$

Now, add equation (4) to equation (3) to eliminate the $y$ terms:

$$\begin{split} (\simplify[!collectNumbers]{{a*abs(b1)} +{sgn*a1*abs(b)}}) x &\,= \simplify[!collectNumbers]{{c*abs(b1)}+{sgn*c1*abs(b)}} \\  \simplify{{a*abs(b1)+sgn*a1*abs(b)} x} &\,= \simplify{{c*abs(b1)+sgn*c1*abs(b)}} .\end{split}$$

So the solution for $x$ is $$x=\simplify{{c*abs(b1)+sgn*c1*abs(b)}/{a*abs(b1)+sgn*a1*abs(b)}}.$$

To obtain a solution for $y$ we can substitute this value of $x$ into either of our initial equations. Using equation (1), we obtain

$$\begin{split} \simplify[noLeadingminus,fractionNumbers,unitFactor]{{a}  {xsimp} + {b}y} &\,=\var{c} \\ \var{b}y &\,= \simplify[!collectNumbers,fractionNumbers]{{c}-{a*xsimp}} \\\var{b}y &\,= \simplify[fractionNumbers]{{c-a*xsimp}} \\y &\,= \simplify[fractionNumbers]{({c/b-a*xsimp/b})} \end{split}$$