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djyliett [7]
3 years ago
15

Find three consecutive odd integers such that six times the second decreased by twice the first is equal to twenty more than the

sum of the second and third
Mathematics
1 answer:
Kitty [74]3 years ago
6 0
Let
x--------> the first <span>odd integer
x+2-----> </span>the second odd integer
x+4-----> the third odd integer

we know that
6(x+2)-2x=20+(x+2)+(x+4)
6x+12-2x=26+2x
4x-2x=26-12
2x=14
x=7

the answer is 
the numbers are
7, 9 and 11
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