[tex] \frac{7}{8} } < \frac{2n-1}{24} < \frac{4}{3} [/tex]
Rezolv prima data [tex] \frac{7}{8} < \frac{2n-1}{24} [/tex] (inmultesc ecuatia cu 8, pentru a se reduce)
[tex] \frac{7}{8} < \frac{2n-1}{24} / (*8) => 7 < \frac{2n-1}{3} [/tex]
21 < 2n-1
21+1 < 2n
22<2n (impart totul la 2)
11<n
apoi rezolv a doua parte : [tex] \frac{2n-1}{24} < \frac{4}{3} [/tex] (inmultesc cu 3)
[tex] \frac{2n-1}{8} <4 [/tex] (inmultesc cu 8)
[tex]2n-1 < 32[/tex]
2n <33
n < [tex] \frac{33}{2} [/tex]
In final am obtinut ca n > 11 si n < [tex] \frac{33}{2} [/tex]
Raspuns final : n ∈ (11, [tex] \frac{33}{2} [/tex] )