Question

a.) Determinati n∈ IN* pentru care $\frac{23}{3n+2}$ este echiunitara .
b.)Determinati n∈ IN* care verifica inegalitatile :$\frac{1}{3}$ < $\frac{n}{6}$ < $\frac{8}{9}$

Answer 1

a) $\frac{23}{3n+2}$ echiunitara => 3n+2=23 => 3n=21, n=7.
b) Aducem fractiile la acelasi numitor: adica 18. Avem:
6/18<3n/18<16/18, deci $\frac{6}{18} < \frac{3n}{18} < \frac{16}{18}$ ⇒6<3n<18 => n∈{3,4,5}

Answer 2

 
 a)  23 = 3n + 2
3n = 23 - 2
n = 21 / 3 = 7
 
      
$\frac{1}{3} < \frac{n}{6} < \frac{8}{9} \\ \text{aducem fractiile la acelasi numitor} \\ \frac{6}{18} < \frac{3n}{18} < \frac{16}{18} \\ \\ 6 <3n<16 \\ n>6/3 =>n>2 \\ n<16/3 => n<5,(3) \\ => Solutia: \\ n \;apartine\;\{3;4;5\}$