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Solution: Let x =# of Partridges, y = # of Pigeons, z= # of Sparrows. Then x+y+z=30 30x+20y+5z=300 (Simplified: 6x+4y+z=60) 6x+4y+z=60 ‐ (x+ y+z=30) 5x+3y = 30 3y=30‐5x Y = 10‐5/3x As he bought whole birds and had a mix, we can assume x, y, and z are all integers greater than 0. If x and y are both integers, x must be a multiple of 3. [y = 10 ‐ (5/3)x] As y>0, x must satisfy 0<x<6. Therefore, x = 3. Plugging in for the rest gives y=5, z=22.