document.write( "Question 133852: The sum of the digits of a two-digit number is 13. If the digits are reversed, the new number is 4 less than twice the original number. Find the original number.\r
\n" ); document.write( "\n" ); document.write( "I an very confused with this problem.\r
\n" ); document.write( "\n" ); document.write( "It's be greatly appreciated if I could get help!
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Algebra.Com's Answer #97894 by solver91311(24713)\"\" \"About 
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10s digit: x
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\n" ); document.write( "\n" ); document.write( "The original number: \"n%5Bo%5D=10x%2By\", because a number like 24 could be represented as \"2%2A10%2B4\"\r
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\n" ); document.write( "\n" ); document.write( "Reversing the digits makes a new number: \"n%5Bn%5D=10y%2Bx\"\r
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\n" ); document.write( "\n" ); document.write( "Twice the original number is \"2%2810x%2By%29\" and 4 less than that is \"2%2810x%2By%29-4\" which we know to be equal to the new number, so:\r
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\n" ); document.write( "\n" ); document.write( "\"10y%2Bx=2%2810x%2By%29-4\"\r
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\n" ); document.write( "\n" ); document.write( "Simplify:
\n" ); document.write( "\"10y%2Bx=20x%2B2y-4\"
\n" ); document.write( "\"10y-2y%2Bx-20x=-4\"
\n" ); document.write( "\"8y-19x=-4\"\r
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\n" ); document.write( "\n" ); document.write( "Since we know that the sum of the digits is 13, we can say \"x%2By=13\", or in other words, \"y=13-x\"\r
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\n" ); document.write( "\n" ); document.write( "Substitute:
\n" ); document.write( "\"8%2813-x%29-19x=-4\"
\n" ); document.write( "\"104-8x-19x=-4\"
\n" ); document.write( "\"-27x=-108\"
\n" ); document.write( "\"x=4\"\r
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\n" ); document.write( "\n" ); document.write( "So the 10s digit of the original number (and the ones digit of the new number) is 4. Therefore the ones digit of the original number must be \"13-4=9\", and the original number must be 49.\r
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\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "Number is: 49
\n" ); document.write( "\"4%2B9=13\"\r
\n" ); document.write( "\n" ); document.write( "\"2%2A49-4=98-4=94\" which is the original number with the digits reversed.
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