Răspuns:
Explicație pas cu pas:
a) 3(x-1) ≤ 2(x-3)+7 <=>
3x-3 ≤ 2x-6+7 <=>
3x-2x ≤ 1+3 => x ≤ 4 =>
x ∈ {0 ; 1 ; 2 ; 3 ; 4}
b) 2x+9-3(x-2) ≤ 4(3-x)+18 <=>
2x+9-3x+6 ≤ 12-4x+18 <=>
-x+15 ≤ 30-4x =>
4x-x ≤ 30-15 =>
3x ≤ 15 I:3 => x ≤ 5 =>
x ∈ {0 ; 1 ; 2 ; 3 ; 4 ; 5}
c) 3(4-3x) + 2(5x - 3) ≤ 12-x <=>
12-9x+10x-6 ≤ 12-x <=>
x+6 ≤ 12-x <=>
x+x ≤ 12-6 =>
2x ≤ 6 I:2 => x ≤ 3 =>
x ∈ {0 ; 1 ; 2 ; 3}
d) 3(2x-5)-4(1-2x) ≤ 6(2x + 3)-25 <=>
6x-15-4+8x ≤ 12x+18-25 <=>
14x-19 ≤ 12x -7 =>
14x-12x ≤ 19-7 =>
2x ≤ 12 I:2 => x ≤ 6 =>
x ∈ x ∈ {0 ; 1 ; 2 ; 3 ; 4 ; 5 ; 6}