document.write( "Question 179292: The sum of 4 times Joan's age and 3 times Jim's age is 47. Jim is 1 year less than twice as old as Joan. Find each of their ages. \n" ); document.write( "
Algebra.Com's Answer #134327 by solver91311(24713)![]() ![]() You can put this solution on YOUR website! \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let Joan's age be represented by x. Then four times Joan's age would be 4x.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let Jim's age be represented by y. Then three times Jim's age would be 3y.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Four times Joan's age (4x), the sum of (+), three times Jim's age (3y) is (=) 47. So: \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Jim is (y =) one year less than twice as old as Joan (2x - 1). So: \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Substitute 2x - 1 for y in \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Check:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "John \n" ); document.write( " \n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |