Question 883636: Find two consecutive odd interegers such that the smaller one is ten more than one-fifth the larger
Answer by algebrahouse.com(1659)   (Show Source):
You can put this solution on YOUR website! x = larger
x/5 + 10 = smaller {the smaller one is ten more than one-fifth the larger}
Consecutive odd integers increase by two from one integer to the next integer
If the larger integer is x, the smaller integer could also be represented as x - 2
x - 2 = x/5 + 10 {the smaller, x - 2, ten more than one-fifth the larger}
5x - 10 = x + 50 {multiplied entire equation by 5 to eliminate fractions}
4x = 60 {added 10 and subtracted x from each side}
x = 15 {divided each side by 4}
x/5 + 10 = 13 {substituted 15, in for x, into x/5 + 10}
13 and 15 are the two consecutive odd integers
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