Question

clasa a 4-a
suma a 3 nr   x,y,z  este 312,daca din primul s-ar scadea 2/5 din al doilea 1/7 iar din al treilea 1/3 cele 3 nr ar fi egale care sunt cele 3 nr?

Answer 1

X+Y+Z=312
X-2/5X=Y-1/7Y=Z-1/3Z
(X-2/5X=Y-1/7Y) factor comun 35( se inmultesc numitorii 5x7) 
35X-14X=35Y-5Y
21X=30Y
X=30/21Y (se simplifica cu 3)
X=10/7Y
Y-1/7Y=Z-1/3Z factor comun 21( se inmultesc numitorii 7x3)
21Y-3Y=21Z-7Z
18Y=14Z 
Z=18/14Y(se simplifica cu 2)
Z=9/7Y
se inlocuiesc valorile lui X si Z in ecuatia de sus(initiala)
10/7Y+Y+9/7Y=312 se da factor comun 7 (se inmulteste cu 7)
10Y+7Y+9Y=312x7
26Y=2184
Y=2184:26
Y=84,atunci X=10/7Y=10/7x84=120
X=120
Z=9/7Y=9/7x84=108
Z=108
120+84+120=312
daca scazi 2/5 din 120, 2/5x120=48,120-48=72
daca scazi 1/7din 84, 1/7x84=12,84-12=72
daca scazi 1/3din 108,1/3x108=36, 108-36=72
cele 3 numere sunt egale 72=72=72

Answer 2

a-2/5=b-1/7=c-1/3=104
a-2/5=104
5a/5-2/5=520/5
5a/5=522/5
a=104,4
105a/105-42/105=105b/105-15/105=105c-35/105=10920/105
105a-42=105b-15=105c-35=10920
105a=105b=105c=11012