SOLUTION: Find four consecutive even integers such that three quarters of the second integral is two less than two thirds of the fourth integer

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Question 1130141: Find four consecutive even integers such that three quarters of the second integral is two less than two thirds of the fourth integer
Found 2 solutions by josmiceli, josgarithmetic:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the consecutive even integers be
n, n+ 2, n + 4, n + 6
+%283%2F4%29%2A%28+n+%2B+2+%29+=+%282%2F3%29%2A%28+n+%2B+6+%29+-+2+
Multiply both sides by +12+
+9%2A%28+n+%2B+2+%29+=+8%2A%28+n+%2B+6+%29+-+24+
+9n+%2B+18+=+8n+%2B+48+-+24+
+n+=+48+-+24+-+18+
+n+=+6+
----------------
The consecutive even integers are:
6, 8, 10, and 12
---------------------
check:
+%283%2F4%29%2A%28+n+%2B+2+%29+=+%282%2F3%29%2A%28+n+%2B+6+%29+-+2+
+%283%2F4%29%2A8+=+%282%2F3%29%2A12+-+2+
+6+=+8+-+2+
+6+=+6+
OK

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
6, 8, 10, 12


(not the only method)
Second and fourth integers are mentioned.
n-4, n-2, n, n+2;
%283%2F4%29%28n-2%29=%282%2F3%29%28n%2B2%29-2
LCD 12;
9%28n-2%29=8%28n%2B2%29-24
n=18%2B16-24
n=10
-
second integer, 8
third integer, 10
fourth integer 12