SOLUTION: Find two consecutive integers such that four times the larger exceeds three times the smaller by 23.

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Question 175156: Find two consecutive integers such that four times the larger exceeds three times the smaller by 23.
Found 4 solutions by Mathtut, stanbon, josmiceli, actuary:
Answer by Mathtut(3670) About Me  (Show Source):
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let the integers be a and b=a+1:
4(a+1)=3a+23
:
4a+4=3a+23
:
highlight%28a=19%29
highlight%28b=a%2B1=20%29
:

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find two consecutive integers such that four times the larger exceeds three times the smaller by 23.
-----------------------
1st "x"
2nd "x+1"
---------------------------
Equation:
4(x+1) - 3x = 23
x = 19 (1st)
x+1 = 20 (2nd)
============================
Cheers,
Stan H.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the smaller integer n
The next larger integer is n%2B1
4%28n%2B1%29+=+3n+%2B+23
4n+%2B+4+=+3n+%2B+23
n+=+19
n%2B1+=+20
The numbers are 19 and 20
check:
4%28n%2B1%29+=+3n+%2B+23
4%2A20+=+3%2A19+%2B+23
80+=+57+%2B+23
80+=+80
OK

Answer by actuary(112) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the first integer.
So the second integer is x+1
The description of the problems leads to the inequality
4*(x+1)>3*x+23 Since x+1 is greater than x
Simplifyling, we have
4x+4>3x+23
4x-3x>23-4
x>19
So x = 20 and x = 21 works