Answer :
[tex]\left\{\begin{array}{ccc}-6x-6y=6&|divide\ both\ sides\ by\ 6\\12x+12y=-12&|divide\ both\ sides\ by\ 12\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-x-y=1\\x+y=-1\end{array}\right}\ \ \ |add\ both\ sides\ of\ the\ equations\\.\ \ \ \ \ \ \ \ 0=0\leftarrow TRUE\\\\therefore\\\\
infinitely\ many\ solutions\Rightarrow \left\{\begin{array}{ccc}y=-x-1\\x\in\mathbb{R}\end{array}\right[/tex]
-6x-6y=6
12x+12x=-12
multiply first equaltion by 2
-12x-12y=12
add to second equation
12x-12x+12y-12y=12-12
0=0
this means that they are the same equation and you really have one equation, but modified
an easy way to find the solution is to make one side equal to one of the placeholders exg y=x+930
12x+12y=-12
divide both sdides by 12
x+y=-1
subtrac x from both sdies
y=-x-1
subsitute values for x and get values for y exg
if x=1 then y=-2
if x=-1 then y=0
if x=0 then y=-1
12x+12x=-12
multiply first equaltion by 2
-12x-12y=12
add to second equation
12x-12x+12y-12y=12-12
0=0
this means that they are the same equation and you really have one equation, but modified
an easy way to find the solution is to make one side equal to one of the placeholders exg y=x+930
12x+12y=-12
divide both sdides by 12
x+y=-1
subtrac x from both sdies
y=-x-1
subsitute values for x and get values for y exg
if x=1 then y=-2
if x=-1 then y=0
if x=0 then y=-1