Question

Determinaţi numerele naturale a si b pentru care a ori b plus 5 ori a plus 3 ori b=15

Answer 1

ab+5a+3b=15
daca a=0, b=5
daca a=1, b+5+3b=15, 4b=10 imposibil
daca a=2, 2b+10+3b=15, 5b=5, b=1

O zi buna!

Answer 2

   
[tex]\displaystyle \\ ab + 5a + 3b = 15 \\ a(b + 5) = -3b + 15 \\ \\ a = \frac{-3b + 15}{b + 5} \\ \\ a = \frac{-3b+3b-3b + 15}{b + 5} \\ \\ a = \frac{-6b+3b + 15}{b + 5} \\ \\ a = \frac{-6b}{b + 5} +\frac{3b + 15}{b + 5} \\ \\ a = \frac{-6b}{b + 5} +\frac{3(b + 5)}{b + 5} \\ \\ a = \frac{-6b}{b + 5} +\frac{3(b + 5)}{b + 5} \\ \\ a = \frac{-6b}{b + 5} +3 [/tex]


[tex]\displaystyle \\ a = 3-\frac{6b}{b + 5} \\ \text{Cautam divizorii naturali ai lui (6).} \\ D_{6} = \{1;~2;~3;~6\} \\ b+5 = 1~~\Longrightarrow~~b=1-5=-4 \notin N \\ b+5 = 2~~\Longrightarrow~~b=2-5=-3 \notin N \\ b+5 = 3~~\Longrightarrow~~b=3-5=-2 \notin N \\ b+5 = 6~~\Longrightarrow~~b=6-5=1 \in N \\ \\ b=\boxed{1} \\ a = 3-\frac{6b}{b + 5} =3-\frac{6\times 1}{1 + 5}=3-\frac{6}{6}=3-1 =\boxed{2} \\ \\ \texttt{Solutia: } \boxed{a= 2~~\texttt{si}~~b=1} [/tex]