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
1 answer:
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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