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3a-b = 29
3a= 29+b
a = (29-b)/3
2[(29-b)/3] + 3b = 45
2(29/3 - b/3) + 3b = 45
58/3 - 2b/3 + 9b/3 = 45
7b/3 = (135/3) - (58/3)
7b = 135-58 = 77
b = 77/7 = 11
a = (29-b)/3
a = (29-11)/3 = 18/3 = 6
Comprobación:
(2*6)+(3*11) = 45
12+33 = 45
Respuesta:
un número o primer número = 6
otro o segundo número = 11