Question 27537: The greater of two consecutive even integers is 6 less than three times the lesser. Find the integers. Found 2 solutions by sdmmadam@yahoo.com, yougan aungamuthu:Answer by sdmmadam@yahoo.com(530) (Show Source):
You can put this solution on YOUR website! The greater of two consecutive even integers is 6 less than three times the lesser. Find the integers.
Consecutive even integers differ by 2
Let the required two consequtive even integers be a and a+2
The greater of a and (a+2) ia 6 less than 3 times the lesser.
That is (a+2)is 6 less than 3a
That is 3a - (a+2) = 6
3a - a -2 = 6
(3a-a) = 6 +2
2a = 8
a = 4
The required consecutive even numbers are 4 and 6
Verification:
The greater should be 6 less than 3times the smaller.
3 times the smaller is 3a = 12
And 6 is of course 6 less than 12
Hence our answer is right
You can put this solution on YOUR website! Lets write this out in english first
Basically we are told that the bigger integer is 6 less than 3
times the smaller integer. this means that when we subtract 6
from three times the smaller integer we get the bigger integer.
Therefore,
3(smaller integer) - 6 = bigger integer
Let us call the first even integer 2k where k can be any integer (the 2 is necessary to emphasise we are dealing with even numbers since any even number must have a factor of 2).
Since we are dealing with consecutive even integers the bigger number will be 2k+2 (we add 2 since consecutive even numbers differ by 2)
Now we use 3(smaller integer) - 6 = bigger integer
3(2k) - 6 = 2k+2
6k-6=2k+2
4k=8
k=2 which means the numbers are 2(2) and 2(2)+2
i.e 4 and 6
Note: the first number is 2k but k = 2 therefore the number 2(2)
the second number is 2k+2 but k=2 therefore the number is 2(2)+2
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