Question 117405This question is from textbook Holt Algebra 1
: Find 3 consecutive integers such that 4 times the middle integer plus the largest integer is 9 less than 6 times the smallest.
This question is from textbook Holt Algebra 1
Answer by psbhowmick(878) (Show Source):
You can put this solution on YOUR website! Let the integers be (n-1), n, (n+1).
4 times middle integer = 4n
4 times middle integer plus the larges integer = 4n + (n-1) = 5n-1
6 times the smallest integer = 6(n-1)
According to the given conditions
5n-1 = 6(n-1) - 9
5n - 1 = 6n - 6 - 9
5n - 1 = 6n - 15
6n - 5n = 15 + 1
n = 16
Hence, the nos. are (16 - 1 =) 15, 16 and (16 + 1 =) 17.
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