|
Question 660308: In a two-digit number. the tens digit is 2 more than twice the ones digit. if the digits are reversed and the new number is doubled, the result is 7 less than the original number. What was the original number?
Answer by kevwill(135)   (Show Source):
You can put this solution on YOUR website! Let x by the tens digit in the number, and y be the ones digit. Then we have
x = 2*y+2 (the tens digit is 2 more than twice the ones digit), and
2(10*y+x) = (10*x+y) - 7
Expanding the second equation:
20*y + 2*x = 10*x + y - 7
20*y + 2*x - 2*x - y = 10*x + y - 7 - 2*x - y
19*y = 8*x - 7
Substituting 2*y+2 for x (from the first equation):
19*y = 8*(2*y+2) - 7
19*y = 16*y + 16 - 7
19*y - 16*y + 9
3*y = 9
y = 3
x = 2*3 + 2 = 8
So the number we're looking for is 83.
Reversing the digits give 38; doubling that gives 76, which is 7 less than the original number, 83.
|
|
|
| |
