algebra
posted by Selenee .
Five times the smallest of three consecutive odd integers is ten more than twice the longest. Find the integers.
The sum of three integers is one hundred twenty three more than the first number. The second number is two more than the first number. the third number is five less than the product of three and the first number. Find the three numbers.
The sum of three consecutive integers is one hundred twenty nine. Find the integers.
The sum of three numbers is forty two. The second number is twice the first number, and the third number is three less than the second number. Find the three numbers.

Five times the smallest of three consecutive odd integers is ten more than twice the longest.
Let the middle integer be x.
then the three odd integers are x2, x, x+2.
Five times the smallest = 5(x2)
ten more than twice the longest= 2(x+2)+10
so
5(x2) = 2(x+2) + 10
5x10 = 2x+4 + 10
5x2x = 4 + 10 + 10
3x = 24
x = 8
The three integers are 6,8 and 10.
You can solve the other problems in the same way. If you have difficulties, post what you have done.
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