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The first question on page 48 of the sixth mathematics classroom exercise book of People's Education Edition.
Suppose there are chickens in the cage, and the number of feet is (14), that is, (6) less than 20 feet; If 1 chicken is replaced by 1 rabbit, the number of feet will increase by (2), that is, (3) chickens will be replaced by (3) rabbits. Therefore, there are (4) chickens and (3) rabbits in the cage.

You can also assume that there are rabbits in the cage, then (the number of feet is only 28, and more than 20 feet are 8); If 1 rabbit is replaced by 1 chicken, the number of feet will be reduced by (2), that is, (4) rabbits will be replaced by (4) chickens. Therefore, there are (4) chickens and (3) rabbits in the cage.

Solution: Suppose there are X chickens in the cage. Or solution: suppose there are x rabbits in the cage.

2X+4 *(7 X)= 20 4X+2 *(7 X)= 20

X=4 X=3

Find the number of rabbits again: (7)-(4)=3 (only) Find the number of rabbits again: (7)-(3)=4 (only)

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