Question 187388
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Given{{{system(2x+3y= 12(EQN1),x+y=4(EQN2)) }}}

In EQN 2 we get--->{{{y=4-x}}}, EQN 3, which we subst. in EQN 1:
{{{2x+3(highlight(4-x))=12}}}
{{{2x+12-3x=12}}}---->{{{3x-2x=12-12}}}---->{{{red(x=0)}}}
Substitute x=0 in EQN 3:
{{{y=4-0}}}----->{{{red(y=4)}}}
There you go, Point of Intersection (0,4).
See graph to verify:
{{{drawing(400,400,-8,8,-8,8,grid(1),graph(400,400,-8,8,-8,8,4-(2/3)x,4-x),blue(circle(0,4,.12)))}}}---> RED Line (EQN 1); GREEN Line (EQN 2)
Thank you,
Jojo</font>