document.write( "Question 115477This question is from textbook algebra 1
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document.write( ": A two-digit number is five times its units digit. if the digits are reversed the resulting number is 27 more than the original number, find the original number? \n" );
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Algebra.Com's Answer #84041 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! Let x = 10's digit and y = the units digit \n" ); document.write( ": \n" ); document.write( "Write an equation for what it says, simplify as much as you can: \n" ); document.write( "\"A two-digit number is five times its units digit.\" \n" ); document.write( "10x + y = 5y \n" ); document.write( "10x = 5y - y \n" ); document.write( "10x = 4y \n" ); document.write( "x = \n" ); document.write( "x = .4y \n" ); document.write( ": \n" ); document.write( "\"if the digits are reversed the resulting number is 27 more than the original number\" \n" ); document.write( "10y + x = 10x + y + 27 \n" ); document.write( ": \n" ); document.write( "10y - y = 10x - x + 27 \n" ); document.write( ": \n" ); document.write( "9y = 9x + 27 \n" ); document.write( ": \n" ); document.write( "Simplify, divide by 9: \n" ); document.write( "y - x + 3 \n" ); document.write( ": \n" ); document.write( "Substitute .4y for x in the above equation, find y \n" ); document.write( "y = .4y + 3 \n" ); document.write( "y - .4y = 3 \n" ); document.write( ".6y = 3 \n" ); document.write( "y = 3/.6 \n" ); document.write( "y = 5 \n" ); document.write( ": \n" ); document.write( "x = .4y \n" ); document.write( "x = .4(5) \n" ); document.write( "x = 2 \n" ); document.write( ": \n" ); document.write( "Our original two digit number is 25: \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check our solution in the statement: \n" ); document.write( " \"if the digits are reversed the resulting number is 27 more than the original number, \" \n" ); document.write( "52 = 25 + 27; confirms our solution \n" ); document.write( ": \n" ); document.write( "Did this make sense to you? Not that hard, right? \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |