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Jamon = j
2q + 1j = $140
1q + 3j = $270
Usamos metodo de eliminacion y multiplicamos la ecuacion 2 por "-2"
-2(1q + 3j = $270) = -2q - 6j = -540
El sistema queda asi y resolvemos las restas y sumas correspondientes
2q + 1j = 140
-2q - 6j = -540
0 - 5j = -400
-5j = -400
j = -400/-5
j = 80
Resultado
1k de jamon cuesta 80