[tex]\bf \begin{cases}
12x+3y=3\\
y=4x+1
\end{cases}\\\\
-------------------------------\\\\
12x+3y=3\implies 3y=-12x+3\implies y=\cfrac{-12x+3}{3}
\\\\\\
y=\cfrac{-12}{3}x+\cfrac{3}{3}\implies \boxed{y=-4x+1}[/tex]
now, recall that the slope-intercept form for the equation is y = mx + b.
now, the second equation is already in slope-intercept form, so we don't have to solve it for "y". Now, what's their slopes anyway?
[tex]\bf y=\stackrel{slope}{-4}x\stackrel{y-intercept}{+1}\qquad \qquad\qquad \qquad y=\stackrel{slope}{4}x\stackrel{y-intercept}{+1}[/tex]
notice, one has a slope of -4, and the other has a slope of 4.
since the slopes differ, the graph of the equations aren't parallel, parallel lines have the same slope. They don't coincide either, coinciding lines have exactly the same equation.
Since their slopes differ, they do intersect at some point.
