The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger. What are the numbers?

1 Answer
smendyka
Nov 11, 2016

#v >= 3# and #v+1 >= 4#

Explanation:

First let's get or define our variables. We can call the first variable #v#. Then because the problem states they are two consecutive integers which means the second integer is one more the first integer the second integer can be called #v + 1#.

"Nine times the smaller" can be written as #9v# and "5 times the larger" can be written as #5(v + 1)#.

"is at most" means we have an inequality and specifically a #<=# or less than or equal to inequality.

So, to write the inequality for the entire problem we have:

#v + (v + 1) <= 9v - 5(v + 1)#

Expanding the terms in parenthesis and then grouping like terms on each side of the inequality gives:

#v + v + 1 <= 9v - 5v - 5#

#2v + 1 <= 4v - 5#

Next we solve for #v# while keeping the inequality balanced:

#2v + 1 - 2v + 5 <= 4v - 5 - 2v + 5#

#6 <= 2v#

#6/2 <= (2v)/2#

#3 <= v#

To "flip" the inequality so the #v# is on the left side we have to "flip " the inequality so #<=# becomes #>=# and the inequality can be written as:

#v >= 3#

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