Question 935586
{{{-2x+3y>= -12}}}.............(1) 
{{{x+y<3}}}...............(2)
__________________________________solve both for {{{y}}}

{{{y>= (2/3)x-4}}}.............(1) 
{{{y<-x+3}}}...............(2)

graph the first inequality:

{{{y>= (2/3)x-4}}}.............(1) 

you need two points to draw a graph:

if {{{x=0}}} then {{{y>=-4}}}; so, first point is ({{{0}}},{{{-4}}})

if {{{y=0}}} then {{{0>=(2/3)x-4}}}=>{{{4>=(2/3)x}}}=>{{{4/(2/3)>=x}}}=>{{{6>=x}}}; so, second point is ({{{6}}},{{{0}}})

{{{drawing( 600,600, -10, 10, -10, 10,circle(6,0,.14),circle(0,-4,.14), graph( 600,600, -10, 10, -10, 10, (2/3)x-4)) }}} 

since we have >=,  line is part of solution
so, shade a part above a line; that is solution


{{{drawing( 600,600, -10, 10, -10, 10,circle(6,0,.14),circle(0,-4,.14), graph( 600,600, -10, 10, -10, 10, y>= (2/3)x-4)) }}} 




graph second inequality on the same graph


{{{y<-x+3}}}...............(2)

if {{{x=0}}} then {{{y<3}}}; so,we take {{{y=3}}} first point is ({{{0}}},{{{3}}})

if {{{y=0}}} then {{{x<3}}}; so, second point is ({{{3}}},{{{0}}})



{{{drawing( 600,600, -10, 10, -10, 10,circle(0,3,.14),circle(3,0,.14),circle(6,0,.14),circle(0,-4,.14), graph( 600,600, -10, 10, -10, 10, (2/3)x-4,-x+3)) }}}

since given that {{{y}}} is less then-our line is not part of solution and need to be dashed line (I can't do it here)

now shade solution part

{{{drawing( 600,600, -10, 10, -10, 10,circle(0,3,.14),circle(3,0,.14),circle(6,0,.14),circle(0,-4,.14), graph( 600,600, -10, 10, -10, 10, (2/3)x-4,y<-x+3)) }}}


finally, we can shade solution to the system:


{{{drawing( 600,600, -10, 10, -10, 10,locate(-7,1,"SHADE_BETWEEN_THE_LINES"),locate(-7,-2,"up_to_intersection_point"),circle(0,3,.14),circle(3,0,.14),circle(6,0,.14),circle(0,-4,.14), graph( 600,600, -10, 10, -10, 10, (2/3)x-4,-x+3)) }}}