SOLUTION: A number consists of two digits such that the digit in ten's place is less by 2 than the digit in the units place. Three times the number added to 6/7 times the number obtained b

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Question 935490: A number consists of two digits such that the digit in ten's place is less by 2 than the digit in the units place. Three times the number added to 6/7 times the number obtained by reserving the digits equals 108. the sum of the digits in the number is:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the tens digit = +t+
Let the units digit = +u+
The value of the number is:
+10t+%2B+u+
-------------------------
(1) +t+=+u+-+2+
(2) +3%2A%28+10t+%2B+u+%29+%2B+%28+6%2F7+%29%2A%28+10u+%2B+t+%29+=+108+
------------------------------------------
(2) +30t+%2B+3u+%2B+%28+60%2F7+%29%2Au+%2B+%28+6%2F7+%29%2At+=+108+
(2) +210t+%2B+21u+%2B+60u+%2B+6t+=+108%2A7+
(2) +216t+%2B+81u+=+756+
(2) +24t+%2B+9u+=+84+
(2) +8t+%2B+3u+=+28+
and, by substitution:
(2) +8%2A%28+u+-+2+%29+%2B+3u+=+28+
(2) +8u+-+16+%2B+3u+=+28+
(2) +11u+=+44+
(2) +u+=+4+
and, since
(1) +t+=+u+-+2+
(1) +t+=+4+-+2+
(1) +t+=+2+
-------------
The number is +24+
The sum of the digits is +6+
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check answer:
(2) +3%2A%28+10t+%2B+u+%29+%2B+%28+6%2F7+%29%2A%28+10u+%2B+t+%29+=+108+
(2) +3%2A%28+10%2A2+%2B+4+%29+%2B+%28+6%2F7+%29%2A%28+10%2A4+%2B+2+%29+=+108+
(2) +3%2A24+%2B+%28+6%2F7+%29%2A42+=+108+
(2) +72+%2B+36+=+108+
(2) +108+=+108+
OK