Question

$Sa~se~rezolve~in~\mathbb~{Z}*~ecuatia:~ \frac{1}{2x}+ \frac{1}{3y}= \frac{1}{4}.$

Answer 1

[tex] \frac{1}{2x} + \frac{1}{3y} = \frac{1}{4}|\cdot 12xy \\ 6y+4x=3xy\\ 3xy-6y=4x\\ y(3x-6)=4x\\ y= \frac{4x}{3x-6} \in Z=>3y\in Z=>\frac{12x}{3x-6} \in Z=> \frac{4(3x-6)+24}{3x-6} \in Z=>\\ 4+ \frac{8}{x-2} \in Z=>\frac{8}{x-2} \in Z\\ x-2=\pm1;\pm2;\pm4;\pm8\\ x=3=>y=4\\ x=6=>y=2[/tex]