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November 26, 2014

November 26, 2014

Posted by **Tamir** on Friday, August 13, 2010 at 9:41pm.

When the digits of a two-digit number are reversed, the new number is 9 more than the original number, and the sum of the digits of the original number is 11. WHat is the original number?

My answer

Let t = the tens digit of the original number

Let u = the units digit of the original number

10t+u

10u+t

10u+t = 10t+u+9

u+t=11

10u+t = 10t+u+9

10u - u+t - 10t = 9

9u - 9t = 9

u+t = 11

u-t = 1

2u = 12

u = 6

u+t = 11

6+t = 11

t = 5

The original number is 56

Check:

10u+t = 10t+u+9

60+5 = 50+6+9

65 = 65

u+t = 11

6+5 = 11

11 = 11

Thanks in advance for the help. I am just about finished with Algebra, which means I am close to HS graduation.

- Algebra 1 -
**Reiny**, Friday, August 13, 2010 at 9:51pmgood job

In the verification, I would have checked if the answer satisfies the wording of the problem.

- original number 56

- number reversed 65

is this 9 more than the original ? YES

is the sum of the digits of 56 equal to 11? YES

Suppose your equations had been incorrect, but the solution to those incorrect equations were found by correct steps

Substituting the answers would satisfy the incorrect equations, but would not necessarily have been the correct answer.

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