Question

Fie multimea A={x ∈ Z/ ║(o linie numa) x≥-5}.
Determinati A∩S,unde S este multimea sulutiilor pentru urmatoarele inecuati:
a) 2(x+5)+7≤x+20.
b)3(x+4)+9 ∠ x+25
c)x+3∠$\frac{3x+17}{5}$
d) x+2≤$\frac{4x+11}{7}$

Va rog!!

Answer 1

a) [tex]2(x+5)+7 \leq x+20\\ 2x + 10 + 7 \leq x + 20\\ 2x + 17 \leq x+20\\ 2x \leq x + 3\\ x \leq 3[/tex] => x ∈ {3, 2, 1, 0, -1, -2,...}

b) [tex]3(x+4)+9 \ \textless \ x+25\\ 3x + 21 \ \textless \ x + 25\\ 3x \ \textless \ x + 4\\ 2x \ \textless \ 4\\ x \ \textless \ 2[/tex] => x ∈ {1, 0, -1, ...}

c) [tex]x+3 \ \textless \ \frac{3x+17}{5} \\ 5x + 15 \ \textless \ 3x + 17\\ 2x \ \textless \ 2\\ x \ \textless \ 1[/tex] => x ∈ {0, -1, ...}

d) [tex]x+2 \leq \frac{4x+11}{7} \\ 7x + 14 \leq 4x + 11\\ 3x + 3 \leq 0\\ 3x \leq -3\\ x \leq -1[/tex] => x ∈ {-1,-2,-3,...}

A∩S={-1,-2,...}

Answer 2

 A={-5;-4;-3;...+∞};
a) 2x+10+7≤x+20
2x+17≤x+20
x+17≤20
x≤3
S={-∞,...,0,1,2,3}
A∩S={-5,-4,...,3}
b) 3x+12+9<x+25
3x+21<x+25
2x<4
x<2
S={-∞,..,1}
A∩S={-5,-4,...,1}
c)x+3<(3x+17)/5
5x+15<3x+17
2x<2
x<1
S={-∞,...,0}
A∩S={-5,-4,-3,...,0}
d)x+2≤(4x+11)/7
7x+14≤4x+11
3x≤-3
x≤-1
S={-∞,...,-1}
A∩S={-5,-4,...,-1}