<PRE><B>Question 951633</B>

{{{8x-6y=12}}}....eq.1....simplify first, both sides divide by{{{2}}}
{{{12x-9y=18}}}....eq.2...both sides divide by{{{3}}
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{{{4x-3y=6}}}....eq.1
{{{4x-3y=6}}}....eq.2
----------------------------now you got same coefficients for {{{x}}} and {{{x}}} variable which means the two lines are identical, so if you subtract eq.2 from eq.1 you will get 

{{{cross(4x)-3y-cross(4x)-(-3y)=6-6}}}

{{{-3y+3y=0}}}

{{{0=0}}} the two lines are identical if one lies perfectly on top of the other, and intersect at all points of both lines

{{{ graph( 600, 600, -10, 10, -10, 10, 4x/3-2, 4x/3-2) }}} 
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