Question 1153788
first graph given lines (as equalities )

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

he figure formed by each system is a triangle and you can find the coordinates of the vertices so that you find intersection points of the all three lines

given:

{{{y>=-2}}}.......1)-> horizontal line
{{{2x+y<=2}}}.......2)
{{{y<=2x+6}}}.......3)
---------------------------

{{{y>=-2}}}.......1)...........substitute in 2)

{{{2x-2<=2}}}.......2)

{{{2x<=2+2}}}

{{{2x<=4}}}

{{{x<=2 }}}

intersection point of the lines 1) and 2) is at ({{{2}}},{{{-2}}})

{{{y>=-2}}}.......1)...........substitute in 3)

{{{-2<=2x+6}}}.......3) 

{{{-2-6<=2x}}}

{{{-8<=2x}}}

{{{-4<=x}}}


intersection point of the lines 1) and 3) is at ({{{-4}}},{{{-2}}})


{{{2x+y<=2}}}.......2)......solve for {{{y}}}

{{{y<=2-2x}}}.......2).........substitute in 3)

{{{2-2x<=2x+6}}}.......3).....solve for {{{x}}}

{{{2-6<=2x+2x}}}

{{{-4<=4x}}}

{{{-1<=x}}}

then

{{{y<=2-2(-1)}}}

{{{y<=2+2}}}

{{{y<=4}}}

intersection point of the lines 2) and 3) is at ({{{-1}}},{{{4}}})

now, plot these points:


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-1,4,.12),circle(-4,-2,.12),circle(2,-2,.12),
locate(-1,4.5,A(-1,4)),locate(-4,-2.5,B(-4,-2)),locate(2,-2.5,C(2,-2)),

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