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Cantidad de monedas de $100 → p
Entonces:
1 → { m + p = 75
2 → {50m + 100p = 5900
Solución : " A) "
Desarrollo del " SEL 2×2 " :
Despejando "m" de la ecuacion "1"
m + p = 75
m = 75 - p
Reemplazando en "2"
50 ( 75 - p ) + 100p = 5900
3750 - 50p + 100 p = 5900
50p = 5900 - 3750
50p = 2150
p = 2150/50
p = 43
Reemplazando en "1"
m = 75 - p
m = 75 - 43
m = 32
Solucion del " SEL 2×2 " → S={( 32 , 43 )}
Cantidad de monedas de $50 → 32
Cantidad de monedas de $100 → 43