document.write( "Question 113520This question is from textbook Arithmetic and algebra again
\n" ); document.write( ": The sum of the digits of a two-digit number is 7. If the digits are reversed, the new number increased by 3 equals 4 times the original number. Find the original number.\r
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\n" ); document.write( "\n" ); document.write( "here is what I have done:\r
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\n" ); document.write( "original number 10x + y
\n" ); document.write( " new number 10y + x\r
\n" ); document.write( "\n" ); document.write( "x + y = 7\r
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\n" ); document.write( " 10x + y = 10y + x + 3
\n" ); document.write( "10x -x + y -y = 10y -y + x -x + 3
\n" ); document.write( " 9x = 9y + 3
\n" ); document.write( " x = y + 3\r
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\n" ); document.write( " x + y = 7
\n" ); document.write( " (y+3) + y = 7
\n" ); document.write( " 2y + 3 = 7
\n" ); document.write( " 2y +3 -3 = 7 -3
\n" ); document.write( " 2y = 4
\n" ); document.write( " 2y/2 = 4/2
\n" ); document.write( " y = 2\r
\n" ); document.write( "\n" ); document.write( " x + y = 7
\n" ); document.write( " 2 + y = 7
\n" ); document.write( " 2 -2 + y = 7 -2\r
\n" ); document.write( "\n" ); document.write( " y = 5
\n" ); document.write( " x = 2\r
\n" ); document.write( "\n" ); document.write( "I get an answer, but it don't check.
\n" ); document.write( "Every time I do this problem this my answer or fractional. please help
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Algebra.Com's Answer #82620 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!
here is one more solution to this problem...\r
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\n" ); document.write( "\n" ); document.write( "Let ten's digit be \"x\" and unit's digit be \"y\". Then,\r
\n" ); document.write( "\n" ); document.write( "\"Sum\" of the digits = \"x+%2B+y\"\r
\n" ); document.write( "\n" ); document.write( "The number is = \"10x+%2B+y\"\r
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\n" ); document.write( "\n" ); document.write( "\"Sum\" of the digits is \"7\",\r
\n" ); document.write( "\n" ); document.write( "\"x+%2B+y+=+7\"…………………………………..…(1)\r
\n" ); document.write( "\n" ); document.write( "When the digits are reversed, the new number becomes\r
\n" ); document.write( "\n" ); document.write( "= \"10y+%2B+x\"\r
\n" ); document.write( "\n" ); document.write( "New number increased by 3 = 4 times the original number\r
\n" ); document.write( "\n" ); document.write( " (10y + x) + 3 = 4 (10x + y)\r
\n" ); document.write( "\n" ); document.write( "or 10y + x + 3 = 40x + 4y\r
\n" ); document.write( "\n" ); document.write( "\"10y+%2B+x+-+40x+-+4y+=+-3\"\r
\n" ); document.write( "\n" ); document.write( "\"-39x+%2B+6y+=+-3\"\r
\n" ); document.write( "\n" ); document.write( "\"-13x+%2B+2y+=+-1\"…… (Dividing by 3) …(2)\r
\n" ); document.write( "\n" ); document.write( "\"Multiplying\" (1) by 2, we get,\r
\n" ); document.write( "\n" ); document.write( "\"2x+%2B+2y+=+14\"………………………………..…(3)\r
\n" ); document.write( "\n" ); document.write( "\"Subtracting\" (3) from (2),\r
\n" ); document.write( "\n" ); document.write( "\"-13x+%2B+2y+=+-1\"………………..……………..(2)
\n" ); document.write( "\"2x+%2B+2y+=+14\"………………………………..…(3)\r
\n" ); document.write( "\n" ); document.write( "\"-15x+=+-+15\" …….=>……\"x+=+1\"\r
\n" ); document.write( "\n" ); document.write( "Substituting \"x+=+1\" in (1) we will have\r
\n" ); document.write( "\n" ); document.write( "\"1+%2B+y+=+7\" ……=>…\"y+=+6\"\r
\n" ); document.write( "\n" ); document.write( "So, the number is \"10x+%2B+y+=+10+x+1+%2B+6+=+16\"\r
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\n" ); document.write( "When the digits are reversed, the new number is \"%2810y+%2B+x%29+=+61\"\r
\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "\"%2810y+%2B+x%29+%2B+3+=+4+%2810x+%2B+y%29\"\r
\n" ); document.write( "\n" ); document.write( "\"61+%2B+3+=+4+%2A16\"\r
\n" ); document.write( "\n" ); document.write( "\"64+=+64\" \r
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