SOLUTION: two consecutive odd numbers are chosen so that one third of the smaller exceeds one seventh of the larger by 6. find the numbers.

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Question 844234: two consecutive odd numbers are chosen so that one third of the smaller exceeds one seventh of the larger by 6. find the numbers.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the smaller number +n+
The larger number will be +n+%2B+2+
+%281%2F3%29%2An+=+%281%2F7%29%2A%28+n+%2B+2+%29+%2B+6+
-----------------------------
+%281%2F3+%29%2An+=+%28+1%2F7+%29%2An+%2B+2%2F7+%2B+6+
Multiply both sides by +21+
+7n+=+3n+%2B+6+%2B+126+
+4n+=+132+
+n+=+33+
+n+%2B+2+=+35+
The consecutive odd numbers are 33 and 35
check:
+%281%2F3%29%2An+=+%281%2F7%29%2A%28+n+%2B+2+%29+%2B+6+
+%281%2F3%29%2A33+=+%281%2F7%29%2A35+%2B+6+
Multiply both sides by 21
+7%2A33+=+3%2A35+%2B+126+
+231+=+105+%2B+126+
+231+=+231+
OK