document.write( "Question 127974This question is from textbook Structure and Method Book 1
\n" ); document.write( ": 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? \n" ); document.write( "

Algebra.Com's Answer #93809 by ilana(307) $\"\"$ $\"About$

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Let's say the original number is ab, so a is the tens digit and b is the ones digit. That means the reversed number is ba, where b is tens and a is ones. Look at the number 35. 3 is the tens and 5 is the ones, so the total is 3(10)+5(1)=30+5=35. So ba must actually be 10b+a. We know this is 9 more than the original, so 10b+a=10a+b+9. If the digits add to 11, a+b=11.
\n" ); document.write( "Now we have a system of 2 equations: 10b+a=10a+b+9 and a+b=11. Using the first, 10b+a=10a+b+9 becomes 9b=9a+9, divide by 9 and get b=a+1. Plug that b into a+b=11 and get a+(a+1)=11, so 2a+1=11, so 2a=10, and finally, a=5. If a=5, since a+b=11, b=6. So the original number was 56. Check if this is right: 65-56=9, and 5+6=11. Nicely done!:) \n" ); document.write( "

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