Question

1. Daca a+b=40, a*c=527 si b*c=713, atunci a=....., b=......., c=........ . Vreau rezolvarea completa!
2.Un numar natural este cu 35 mai mare decat altul. Impartind suma lor la diferenta, obtinem catul 12 si restul 21. Aflati numerele. Vreau rezolvarea completa!
3.Suma a trei numere naturale este 486. Impartind al doilea numar la primul, se obtine catul 3 si restul 18.Impartind al treilea numar la cel de-al doilea, se obtine catul 2 si restul 12. Sa se afle numerele. 
           Vreau rezolvarea completa! 

Answer 1

a+b=40

a*c=527
b*c=713 adumnam cele doua relatii si obtinem 

a*c+b*c=527+713 dam factor comun pe c 
c*(a+b)=1240
c*40=1240
c=1240:40
c=31

31*b=713
b=713:31
b=23

a*31=527
a=527:31
a=17

==============================================================

b=a-35
(a+b):(a-b)=12 rest 21

a+b=12(a-b) +21
a+a-35=12(a-a-35) +21
2a-35=12x35+21
2a-35=441
2a=441+35
2a=476
a=476:2
a=238

b=238-35
b=203


===============================================================

a+b+c=486
b:a=3 rest 18⇒b=3a+18
c:b=2 rest 12⇒c=2b+12 


c=2(3a+18)+12
c=6a+36+12
c=6a+48

a+3a+18+6a+48=486
10a+66=486
10a=486-66
10a=420
a=420:10
a=42

b=3x42+18
b=144

c=6x42+48
c=300

Verificare: 42+144+300=486

Answer 2

1)ac+bc= c(a+b)

ac+bc= 527 + 713= 1240

c(a+b)=1240

c* 40=1240

c=31

ac=527

a=527:31

a=17

bc=713

c=713:31

c=23

2)a-35=b

(a+b):(a-b)= 12 rest 21

a+b=12(a-b) +21

a+a-35=12(a-a+35) +21

2a-35=12*35 +21

2a-35= 420 +21

2a=441+35

2a= 476

a=238

b=238-35= 203

3)a+b+c=486

b:a=3 rest 18

b=3a +18 

c:b=2 rest 12

c=2b+12

c=2(3a+18)+12

c=6a +36 +12

c=6a +48

a+b+c=486

a+3a+18+6a+48=486

10a +66=486

10a=486-66

10a=420

a=42

b=3*42+18=  126+18= 144

c=6*42 + 48= 252 +48= 300