Question 635812
{{{ 2x+3y=9}}}

{{{3y=-2x+9}}}

{{{y=-2x/3+9/3}}}

{{{y=-(2/3)x+3}}}

now find two points....choose values for {{{x=0}}} and {{{y=0}}}

if {{{x=0}}}, then {{{y=-(2/3)0+3}}}.....{{{y=3}}}

first point is ({{{0}}},{{{3}}})

if {{{y=0}}}, then {{{y=-(2/3)x+3}}}.....{{{-(2/3)x=-3}}}...->..

{{{(2/3)x=3}}}...->..{{{x=3/(2/3)}}}...->..{{{x=3*3/2}}}...->..

{{{x=9/2}}}...->..{{{x=4.5}}}

second point is ({{{4.5}}},{{{0}}})

now graph it


{{{ graph( 600, 600, -10, 15, -10, 10, -(2/3)x+3) }}}