Question

14x+22 pe 2x+1 sa apartina multimii z

Answer 1

(14x +22) /(2x + 1) ⇔  (7x + 11)/(2x+1) ∈ Z ⇒
⇒  (2x+1) | (7x +11)  ⇒  (2x+1) | 2(7x + 11) = 14x + 22    (1)
     (2x+1) | (2x+1)  ⇒ (2x+1) | 7(2x+1) = 14x + 7  (2)
⇒ (2x+1) | [(1) - (2)] = 15 
⇒ (2x+1) ∈ D15    (2x+1) ∈ {-15,-5,-3,-1,1,3,5,15}
2x ∈ {-16,-6,-4,-2,0,2,4,14} 
x ∈ { -8,-2,-1,0,1,2,7}