SOLUTION: The larger of two consecutive integers is 11 more than one half the smaller

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Question 353690: The larger of two consecutive integers is 11 more than one half the smaller
Found 2 solutions by checkley77, jsmallt9:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
x & x+1 are the 2 integers.
x+1=(x+11)/2 cross multiply
2(x+1)=x+11
2x+2=x+11
2x-x=11-2
x=9 ans.
Proof:
9+1=(9+11)/2
10=20/2
10=10

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the larger integer.
Then x-1 would be the next smaller integer.
Now "The larger of two consecutive integers is 11 more than one half the smaller." translates into:
x+=+11+%2B+%281%2F2%29%28x-1%29
Simplifying we get:
x+=+11+%2B+%281%2F2%29x-1%2F2%29
Get rid of the fractions by multiplying both sides by 2:
2x+=+22+%2B+x+-+1
2x+=+21+%2B+x
Subtracting x from each side we get:
x+=+21
"x", as we said earlier, is the larger integer and "x-1" or 20 is the smaller integer. To check you can easily determine that the larger number, 21, is actually 11 more than half of the smaller, 20.