Posted by **Anonymous** on Friday, September 30, 2011 at 12:38am.

The tens digit of a two-digit positive integer is 2 more than three times the ones digit. If the digits are interchanged, the new number 13 less than half the given number. Find the given integer. (Hint: Let x = tens-place digit and y = ones-place digit, then 10x + y is the number.)

- Algebra 2 -
**Peter**, Friday, September 30, 2011 at 2:26am
The tens digit of a two-digit positive integer is 2 more than three times the ones digit. If the digits are interchanged, the new number 13 less than half the given number. Find the given integer. (Hint: Let x = tens-place digit and y = ones-place digit, then 10x + y is the number.)

- Math -
**Peter**, Friday, September 30, 2011 at 2:28am
i think 82 is the answer.

- Math -
**Reiny**, Friday, September 30, 2011 at 7:55am
original:

unit digit: x

tens digit: y

the actual number: 10y + x

number reversed: 10x + y

"the new number 13 less than half the given number" ---> 10x+y < (1/2)(10y+x) by 13

so , take 13 away from the larger part to make them equal.

10x + y = (1/2)(10y+x) - 13

20x + 2y = 10y + x - 26

19x -8y = -26

also y = 3x + 2

use substitution:

19x - 8(3x+2) = -26

19x - 24x - 16 = -26

-5x = -10

x = 2 , then y = 3(2)+2 = 8

the original number is 82

check: number reversed = 28

half the given number = 41

is half the given number (41) less than the number reversed (28) by 13 ??

yes!

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