a) 2x+3=0
2x=-3
x=-3/2
cazul I: daca x∈(-α-3/2)=D1
3[3(-2x-3)+4(x+5)-11]=4(x+9)+36+8x
3[-6x-9+4x+20-11]=4x+36+36+8x
3(-2x)=12x+72
3(-2x)=3(4x-24) :3
-2x=4x-24
-2x-4x=-24-18
-6x=-42 inmultim -1
x=42/6
x=7⇔∉D1
S={-7}
cazul II: x∈[-3/2,+α)=D2
3[3(2x+3)+4(x+5)-11]=4(x+9)+36+8x
3[6x+9+4x+20-11]=4x+36+36+8x
3(10x+18)=12x+72
3(10x+18)=3(4x-24) :3
10x+18=4x-24
10x-4x=-24-18
6x=-42
x=-42/6
x=-7∉D2
x∉Ф
b)2x+5=0
x=-5/2
cazulI: x∈(-α,-5/2)=D1
3[3(-2x-5)+2(x-7)+5]=5(x+2)+x+44
3[-6x-15+2x-14+5]=5x+10+x+44
3(-4x-24)=6x+54
3(-4x-24)=3(2x+18) :3
-4x-24=2x+18
-4x-2x=18+24
-6x=42 inmultim -1
x=-42/6
x=-7∈D1
S1={-7}
cazul II: x∈[-5/2,+α)=D2
3[3(2x+5)+2(x-7)+5)=5(x+2)+x+44
3(6x+15+2x-14+5)=5x+10+x+44
3(8x+6)=6x+54
3(8x+6)=3(2x+18) :3
8x+6=2x+18
8x-2x=18-6
6x=12
x=12/6
x=2∈D2
S2={2}
S={-7,2}
c)2x+7=0
x=-7/2
cazul I: x∈(-α, -7/2)=D1
3(4x-5)-3(-2x-7)+32=6(2x+1)-16
12x-15+6x+21+32=12x+6-16
18x+38=12x-10
18x-12x=-10-38
6x=-48
x=-48/6
x=-8∈D1
S1={-8}
Cazul II: x∈[-7/2,+α)=D2
3(4x-5)-3(2x+7)+32=6(2x+1)-16
12x-15-6x-21+32=12x+6-16
6x-4=12x-10
6x-12x=-10+4
-6x=-6 inmultim -1
x=6/6
x=1∈D2
S2={1}
S={-8,1}
