Question 189589
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Given:{{{system(x=(1/3)y+2(EQN1),-2-y=1(EQN2))}}}
In {{{red(EQN2)}}} we get,
{{{-2-y=1}}}---->{{{-2+1=y}}} ----> {{{y=-1}}}


Subst. {{{y=-1}}} in {{{red(EQN1)}}},
{{{x=(1/3)(highlight(-1))+2=-(1/3)+2=(-2+12)/6=10/6}}}
{{{x=5/3}}}


These two lines intersect at point (5/3,-1).


{{{drawing(400,400,-7,7,-7,7,graph(400,400,-7,7,-7,7,3x-6,-1),blue(circle(5/3,-1,.08)))}}}---> {{{red(RED=EQN1)}}}; {{{green(GREEN=EQN2)}}}


Thank you,
Jojo</font>