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C = moneda de 5
D = meneda de 2
Traduciendo el e unciado
C + D = 151 (1)
5C + 2D = 413 (2)
Resolviendo el sistema
(1) x (- 2)
- 2C - 2C = - 302 (3)
(3) + (2)
3C = 111
C = 111/3
C = 37
C en (1)
37 + D = 151
D = 151 - 37
D = 114
HABÍA 37 MONEDAS DE $5 Y 114 DE $2