document.write( "Question 349919: Roger's age is 1 1/3 times Steven's age. Eight years ago, Roger's age was twice Steven's age. Find Roger's age. \n" ); document.write( "
Algebra.Com's Answer #250090 by mananth(16946) You can put this solution on YOUR website! let steven's age be x \n" ); document.write( "roger's age = 1 1/3 times stevens age. \n" ); document.write( "=4/3 times = 4x/3 \n" ); document.write( "... \n" ); document.write( "8 years ago \n" ); document.write( "steven's age = x-8 \n" ); document.write( "roger's age = 4x/3 - 8 \n" ); document.write( "... \n" ); document.write( "4x/3 -8 = 2(x-8) \n" ); document.write( "(4x-24)/3 = 2x-16 \n" ); document.write( "x3 \n" ); document.write( "4x-24=3(2x-16) \n" ); document.write( "4x-24=6x-48 \n" ); document.write( "6x-4x=48-24 \n" ); document.write( "2x=24 \n" ); document.write( "x=12 steven's age \n" ); document.write( "Roger = 4/3 * 12 \n" ); document.write( "=16 years \n" ); document.write( " |