document.write( "Question 540688: The ages in years of three sisters are consecutive multiples of five. Six years ago the sum of their ages was 72. Find their present ages.\r
\n" ); document.write( "\n" ); document.write( "I have gotten this far:\r
\n" ); document.write( "\n" ); document.write( "Let x= the youngest sisters age now
\n" ); document.write( "x+5 = the middle sisters age now
\n" ); document.write( "xt10 = the oldest sisters age now\r
\n" ); document.write( "\n" ); document.write( "(x-6)+(x-1)+(x+4)=72
\n" ); document.write( "3x+3 =72
\n" ); document.write( "-3+3x+3 =72+-3
\n" ); document.write( "0+3x =69
\n" ); document.write( "3x =69
\n" ); document.write( "1/3×3x =69×1/3
\n" ); document.write( "x =23\r
\n" ); document.write( "\n" ); document.write( "Is this correct?
\n" ); document.write( "Thank you for helping me.
\n" ); document.write( "

Algebra.Com's Answer #353801 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
Let x= the youngest sisters age now
\n" ); document.write( "x+5 = the middle sisters age now
\n" ); document.write( "x+10 = the oldest sisters age now\r
\n" ); document.write( "\n" ); document.write( "(x-6)+(x-1)+(x+4)=72
\n" ); document.write( "3x-3 =72
\n" ); document.write( "-3+3x+3 =72+3
\n" ); document.write( "0+3x =75
\n" ); document.write( "3x =75
\n" ); document.write( "1/3×3x =75×1/3
\n" ); document.write( "x =25\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "25, 30,35 \r
\n" ); document.write( "\n" ); document.write( "CHECK
\n" ); document.write( "6 years ago
\n" ); document.write( "19,24,29
\n" ); document.write( "add up
\n" ); document.write( "72
\n" ); document.write( "
\n" );