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Codrinel008
a fost răspuns

Gaseste numerele naturale a,b,c stiind ca :

a+b=62
a-b=6
4a+11b-3c=393

Răspuns :

a + b = 62
a - b = 6 ⇒ a = 6 + b

6 + b + b = 62
2b = 56
b = 28
a = 6 + 28 = 34

4a + 11b - 3c = 393
4 x 34 + 11 x 28 - 3c = 393
136 + 308 - 3c = 393
-3c = 393 - 136 - 308
-3c = -51
c = 17
HyperCoder
a+b=62
a-b=6
4a+11b-3c=393
a =? ; b = ? ; c = ? 

a + b = 62 => a = 62 - b
a - b = 6 => a = 6 + b
62 - b = 6 + b
b + b = 62 - 6 
2b = 56
b = 56/2
=> b = 28

a + 28 = 62
a = 62 - 28
a = 34

4·34+11·28-3c=393
136 + 308 - 3c = 393
3c = 444 - 393
3c = 51
c = 51/3
=> c = 17

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