Question 703485
{{{y=6}}}....eq.1

{{{3x-2y=-6}}}......eq.2
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{{{3x-2y=-6}}}.....substitute {{{y}}} from eq. 1

{{{3x-2*6=-6}}}

{{{3x-12=-6}}}

{{{3x=-6+12}}}

{{{3x=6}}}

{{{x=6/3}}}

{{{x=2}}}

if {{{x=2}}} then {{{3*2-2y=-6}}}...=>...{{{6-2y=-6}}}...=>...{{{-2y=-6-6}}}

..=>...{{{-2y=-2}}}..=>...{{{y=-12/-2}}}..=>...{{{y=6}}}

so, both lines intersect at ({{{2}}},{{{6}}})


{{{ graph( 600, 600, -10, 10, -10, 10, 6, (3/2)x+3) }}}