Question

Determinati toate perechile de numere întregi (a, b) astfel încât a/7=2/2b+1

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Answer 1

 

$\displaystyle\\\frac{a}{7}=\frac{2}{2b+1}\\\\a(2b+1)=2\times7\\\\a(2b+1)=14\\\\D_{14}=\{-14;~-7;~-2;~-1;~1;~2;~7;~14\}\\\\a(2b+1)=14\\\\2b+1~\text{este numar impar}$

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$\displaystyle\\\text{Perechile de divizori al caror produs este 14 sunt:}\\-14\times(-1)=14\\-2\times(-7)=14\\14\times1=14\\2\times7=14\\------\\\\\text{Solutia 1:}\\\boxed{\bf~a=-14}\\2b+1=-1\implies 2b=-1-1\implies2b=-2\implies b=\frac{-2}{2} \implies \boxed{\bf~b=-1}\\\\\text{Solutia 2:}\\\boxed{\bf~a=-2}\\2b+1=-7\implies 2b=-7-1\implies2b=-8\implies b=\frac{-8}{2} \implies \boxed{\bf~b=-4}$

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$\displaystyle\\\text{Solutia 3:}\\\boxed{\bf~a=14}\\2b+1=1\implies 2b=1-1\implies2b=0\implies b=\frac{0}{2} \implies \boxed{\bf~b=0}\\\\\text{Solutia 4:}\\\boxed{\bf~a=2}\\2b+1=7\implies 2b=7-1\implies2b=6\implies b=\frac{6}{2} \implies \boxed{\bf~b=3}$