document.write( "Question 649980: 1. There’s a two digit number, let’s call it the original number. When we reverse the digits, the new number thus formed, is 27 less than the original number. When 2 is subtracted from the original number, it becomes twice the new number. Find the original number.
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Algebra.Com's Answer #407155 by lwsshak3(11628) $\"\"$ $\"About$

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There’s a two digit number, let’s call it the original number. When we reverse the digits, the new number thus formed, is 27 less than the original number. When 2 is subtracted from the original number, it becomes twice the new number. Find the original number.
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\n" ); document.write( "let u=units digit
\n" ); document.write( "let t=tens digit
\n" ); document.write( "original number=10t+u
\n" ); document.write( "new number=10u+t
\n" ); document.write( "..
\n" ); document.write( "10t+u=10u+t+27
\n" ); document.write( "10t+u-2=2(10u+t)=20u+2t
\n" ); document.write( "..
\n" ); document.write( "9t-9u=27
\n" ); document.write( "8t-19u=2
\n" ); document.write( "..
\n" ); document.write( "3t-3u=9
\n" ); document.write( "8t-19u=2
\n" ); document.write( "..
\n" ); document.write( "24t-24u=72
\n" ); document.write( "24t-57u=6
\n" ); document.write( "33u=66
\n" ); document.write( "u=2
\n" ); document.write( "3t=9+3u=9+6=15
\n" ); document.write( "t=15/3=5
\n" ); document.write( "original number=10t+u=50+2=52
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