document.write( "Question 898326: find two consecutive integers such that the sum of one third of the first integer and one fourth of the second integer is 9. \n" ); document.write( "
Algebra.Com's Answer #544757 by KMST(5328)![]() ![]() You can put this solution on YOUR website! THE ALGEBRA WAY: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "The problem says that \n" ); document.write( " \n" ); document.write( "Multiplying both sides of the equalsign times \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( "The two consecutive integers are \n" ); document.write( " \n" ); document.write( "ANOTHER WAY (Guess and check): \n" ); document.write( "One third of the first integer must be an integer (a whole number), \n" ); document.write( "and one fourth of the second integer must be an integer too. \n" ); document.write( "Otherwise, one or both would be fractions that would add up to a fraction. \n" ); document.write( "The first integer could be 0, 3, 6, 9, 12, 15, 18, 21, 24, ...., \n" ); document.write( "and one third of that would be 0, 1, 2, 3, 4, 5, 6, 7, 8, .... \n" ); document.write( "If we add 1 to each of those choices, the second integer would be 1, 4, 7, 13, 16, 19, 22, 25, ...., but not all of those numbers divide by 4 evenly. \n" ); document.write( "Only 4, and 16 do. \n" ); document.write( "If the second integer were 4 and the first one were 3, \n" ); document.write( "one third of the first integer and one fourth of the second integer would be \n" ); document.write( " \n" ); document.write( "If the second integer were 16 and the first one were 15, \n" ); document.write( "one third of the first integer and one fourth of the second integer would be \n" ); document.write( " |