Question 1000482

{{{x/3-y=2}}}.....eq.1...both sides multiply by {{{3}}}
{{{-x/2+3y/2=-3}}}......eq.2...both sides multiply by {{{2}}}
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{{{x-3y=6}}}.....eq.1
{{{-x+3y=-6}}}
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{{{x-3y-x+3y=6-6}}}

{{{0=0}}}

So we're left with

{{{0=0}}}


which means any {{{x}}} or {{{y}}} value will satisfy the system of equations. So, there are an infinite number of solutions (they coincide with each other) and this system is dependent.