7x= 28 so x is 4 and y is 1.. So (4,1)
Since there are more parakeets than canaries, it is not possible to have only 1 of each bird in each cage <u>and</u> have the same number of birds in each cage.
He could use 42 cages, putting a canary in with the parakeet in 18 of them. Then he would have 18 cages with 2 birds each, and 24 cages with 1 bird each.
The only way to have the same number of birds (1) in all cages is to have 60 cages, 42 of which have 1 parakeet, and 18 of which have 1 canary.
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If more than 1 of each kind of bird can be put in the cage, the collection of birds could be put into 6 cages, each of which would be home to 7 parakeets and 3 canaries.
9514 1404 393
Answer:
no solution
Step-by-step explanation:
The equations are inconsistent. The set of equations reduces to ...
x + 2y + 3z = 12
x + 2y + 3z = 30
x + 2y + 3z = 60
No values of x, y, and z can satisfy all three equations. There is no solution.
