document.write( "Question 883636: Find two consecutive odd interegers such that the smaller one is ten more than one-fifth the larger \n" ); document.write( "
Algebra.Com's Answer #533655 by algebrahouse.com(1659)![]() ![]() You can put this solution on YOUR website! x = larger \n" ); document.write( "x/5 + 10 = smaller {the smaller one is ten more than one-fifth the larger}\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Consecutive odd integers increase by two from one integer to the next integer \n" ); document.write( "If the larger integer is x, the smaller integer could also be represented as x - 2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x - 2 = x/5 + 10 {the smaller, x - 2, ten more than one-fifth the larger} \n" ); document.write( "5x - 10 = x + 50 {multiplied entire equation by 5 to eliminate fractions} \n" ); document.write( "4x = 60 {added 10 and subtracted x from each side} \n" ); document.write( "x = 15 {divided each side by 4} \n" ); document.write( "x/5 + 10 = 13 {substituted 15, in for x, into x/5 + 10}\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "13 and 15 are the two consecutive odd integers \n" ); document.write( " For more help from me, visit: www.algebrahouse.com \n" ); document.write( " |