Find three consecutive odd integers such that the sum of the smaller two is three times larger increased by seven
1 answer:
Answer:
three consecutive odd integers: 2n-1 2n+1 2n+3
that the sum of the smaller two is three
times larger increased by seven: 2n-1 + 2n+1 = 3(2n+3) +7
4n = 6n+ 9 +7
4n = 6n+ 16
4n -6n = 16
-2n = 16
n = 16/(-2)
n=-8
a) 2n -1 = 2(-8) -1 = -17
b) 2n+1 = 2(-8)+1 = -15
c) 2n+1 = 2(-8)+3 = -13
Ans. -17 ; -15 ; -13
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