SOLUTION: Please help. I've been trying to solve this problem: Find two consecutive odd integers whose sum is 2 less than twice the larger. my working equation is: (x)+(x+2)=2(x+2)-2 (x is
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Question 254441: Please help. I've been trying to solve this problem: Find two consecutive odd integers whose sum is 2 less than twice the larger. my working equation is: (x)+(x+2)=2(x+2)-2 (x is an odd integer; x+2 is the larger integer). All I get is 0! There must be something wrong with what I'm doing. Please help. Thanks! Found 2 solutions by palanisamy, richwmiller:Answer by palanisamy(496) (Show Source):
You can put this solution on YOUR website! x+x+2=2(x+2)-2
2x+2=2x+4-2
2x=2x
x=1 or zero
check 1
1+3=2(3)-2
4=6-2
4=4
ok
Integers are 1 and 3
Another way to guarantee only odd answers for the variable is make the variable 2n+1 and 2n+3
2n+1+2n+3=2(2n+3)-2
4n+4=4n+6-2
4n=4n
n=0 or 1
odd must be 2n+1
2*0+1=1
check
1+3=2*3-2
4=4
ok
2*1+1=3
check
3+5=2(5)-2
8=8
ok
So this way we get pairs (1 and 3) and (3 and 5)
But I think ALL odd numbers fit the criteria
23 +25=2(25)-2
48=50-2
(
