Question

1,Suma a trei nr naturale este 108.Daca impartim primul nr. la al II-lea obtinem catul 1 si rest 12,iar daca impartim primul nr. la al III-lea obtinem catul 2.Sa se afle numerele?

2,Suma a trei nr este 195. Sa se afle nr. stiind ca al treilea este dublul primului nr iar al II-lea intrece cu o treime primul nr.
3, Sunt trei nr.Impartim primul nr la al II-lea obtinem catul 1 rest 20, iar impartind primul nr. la al III-lea obtinem catul 1 rest 40 si suma primelor doua nr este 160.Sa se afle nr.

Answer 1

 

$\displaystyle\\1)\\x+y+z=108\\x:y=1~rest~12\\x:z=2\\..........\\x+y+z=108\\x=y\cdot1+12\\x=2z\\..........\\x+y+z=108\\x-y=12\\x-2z=0\\..........~~~\text{Adunam primele 2 ecuatii si scapam de }~y.\\2x+z=120\\x-2z=0~~~~~\implies \boxed{x=2z}~~\text{substitutie.}\\..........\\2\cdot2z+z=120\\2\cdot2z+z=120\\4z+z=120\\5z=120\\\\z=\frac{120}{5}\\\\\boxed{z=24}\\\text{Ne intoarcem la substitutie.}\\x=2z\\x=2\cdot24\\\boxed{x=48}\\x+y+z=108\\48+y+24=108\\y=108-48-24\\\boxed{y=36}$

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$\displaystyle\\2)\\x+y+z=195\\2x=z\\x+\frac{x}{3}=y~~\Big|\cdot3\\..........\\x+y+z=195\\2x-z=0\\3x+x=3y\\..........\\x+y+z=195\\2x-z=0~~~\implies~\boxed{z=2x}~~\text{substitutie}\\4x-3y=0~~~\implies~\boxed{y=\frac{4x}{3}}~~\text{substitutie}\\..........\\x+y+z=195\\\\x+\frac{4x}{3}+2x=195~~\Big|\cdot3\\\\3x+4x+6x=585\\\\13x=585\\\\x=\frac{585}{13}\\\\\boxed{x=45}\\\text{Ne intoarcem la substitutii.}\\\\z=2z\\\boxed{z=90}\\\\y=\frac{4x}{3}=\frac{4\cdot45}{3}==4\cdot15\\\boxed{y=60}$

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$\displaystyle\\3)\\x:y=1~rest~20\\x:z=1~rest~40\\x+y=160\\..........\\x=y+20\\x=z+40\\x+y=160\\..........\\x-y=20\\x-z=40\\x+y=160\\.......... \text{Adunam prima si a treia ecuatie.}\\2x  / = 180\\x=180:2\\\boxed{x=90}\\x-z=40\\90-z=40\\z=90-40\\\boxed{z=50}\\x+y=160\\90+y=160\\y=160-90\\\boxed{y=70}$