Solving system of linear equations with rationals Problem:$$(x+11):(x+6)=(y+12):(y+7)\land(y+1):(x-1)=y:x$$
When I tried using $\frac{a1}{b1}=\frac{a2}{b2}=\frac{k1a1+k2a2}{k1b1+k2b2}$ for $\frac{x+11}{x+6}=\frac{y+12}{y+7}$ where k1 = 1 and k2 = -1, this is what I got:$$\frac{x+11-y-12}{x+6-y-7}=\frac{x-y-1}{x-y-1}=1$$
Which would mean $\frac{x+11}{x+6}=1$ and $\frac{y+12}{y+7}=1$, which are false.  
If I instead transform $(x+11):(x+6)=(y+12):(y+7)$ into $\frac{x+11}{y+12}=\frac{x+6}{y+7}$ then $$\frac{x+11-x-6}{y+12-y-7}=\frac{5}{5}=1$$
Which is possible for $\frac{x+11}{y+12}$ and $\frac{x+6}{y+7}$. The equations are easy and could be solved simply by other means. I must be missing something fundamental.  
Please help. Thanks.

$
\left\{
\begin{array}{l}
 \dfrac{x+11}{x+6}=\dfrac{y+12}{y+7} \\
 \dfrac{y+1}{x-1}=\dfrac{y}{x} \\
\end{array}
\right.
$
$
\left\{
\begin{array}{l}
 (x+11)(y+7)=(x+6)(y+12) \\
 x(y+1)=y(x-1) \\
\end{array}
\right.
$
$
\left\{
\begin{array}{l}
 xy+7x+11y+77=xy+12x+6y+72 \\
 xy+x=xy-y\\
\end{array}
\right.
$
$
\left\{
\begin{array}{l}
 x-y=1 \\
 x+y=0\\
\end{array}
\right.
$
$
\left\{
\begin{array}{l}
 x=\dfrac{1}{2} \\
 y=-\dfrac{1}{2}\\
\end{array}
\right.
$
Hope this can be useful
