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.

Hint:
First absurd result came up as $x-y-1$ is actually $=0$ here.
For example $$\dfrac23=\dfrac46=\dfrac{2\cdot2-4}{2\cdot3-6}=?$$
Just cross multiply to form two simultaneous equations in $x,y$
