document.write( "Question 9360: A two-digit number is such that the product of its digits is 12.When 36 is added to this number, the digits interchange their places.Find the number. \n" ); document.write( "
Algebra.Com's Answer #5072 by longjonsilver(2297)\"\" \"About 
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let the 2-digit number be thought of as xy. This means the number is really 10x+y, as 34 means 30+4--> 3x10 + 4\r
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\n" ); document.write( "\n" ); document.write( "Product is xy = 12 --eqn1\r
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\n" ); document.write( "\n" ); document.write( "straightaway, this means we have either 26 or 34. Looking at these, adding 36 to both gives 62 or 70. So the answer is 26.\r
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\n" ); document.write( "\n" ); document.write( "Anyway, algebraically...\r
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\n" ); document.write( "\n" ); document.write( "adding 36 to the number is 10x+y+36
\n" ); document.write( "and this equals \"yx\" or rather 10y+x\r
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\n" ); document.write( "\n" ); document.write( "so, 10x+y+36 = 10y+x
\n" ); document.write( "--> 9x - 9y = -36
\n" ); document.write( "--> x - y = -4\r
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\n" ); document.write( "\n" ); document.write( "and subbing in eqn1 into this gives \"12%2Fy+-+y+=+-4\"\r
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\n" ); document.write( "\n" ); document.write( "\"12+-+y%5E2+=+-4y\"
\n" ); document.write( "\"y%5E2+-+4y+-+12+=+0\"
\n" ); document.write( "(y+2)(y-6) = 0
\n" ); document.write( "so y = -2 or y=6\r
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\n" ); document.write( "\n" ); document.write( "ignore the -2 value...we are looking for a positive number, so y=6. Hence from eqn1, x = 2...number is 26!\r
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\n" ); document.write( "\n" ); document.write( "jon.
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