Solution 2.2:6d

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At the point where the lines cut each other, we have a point that lies on both lines and which must therefore satisfy the equations of both lines,

\displaystyle x+y+1=0\qquad\text{and}\qquad x=12\,\textrm{.}

We obtain the solution to this system of equations by substituting \displaystyle x=12 into the first equation

\displaystyle 12+y+1=0\quad\Leftrightarrow\quad y=-13\,\textrm{,}

which gives us the point of intersection as (12,-13).