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
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document.write( "I have gotten this far:\r
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document.write( "Let x= the youngest sisters age now
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document.write( "x+5 = the middle sisters age now
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document.write( "xt10 = the oldest sisters age now\r
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document.write( "(x-6)+(x-1)+(x+4)=72
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document.write( "3x+3 =72
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document.write( "-3+3x+3 =72+-3
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document.write( "0+3x =69
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document.write( "3x =69
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document.write( "1/3×3x =69×1/3
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document.write( "x =23\r
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document.write( "Is this correct?
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document.write( "Thank you for helping me. \n" );
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Algebra.Com's Answer #353801 by mananth(16946)![]() ![]() 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( " |