document.write( "Question 46232: Lisa is 1/3 the age of Sally. 6 years ago Lisa was 1/9 the age of Sally. How old is Lisa and How old is Sally? Thank you. Please show step for step. I need help \n" ); document.write( "
Algebra.Com's Answer #30688 by wuwei96815(245)\"\" \"About 
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I found this problem difficult to solve.\r
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\n" ); document.write( "\n" ); document.write( "Let x=Sally's age
\n" ); document.write( "Then 1/3x=Lisa's age\r
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\n" ); document.write( "\n" ); document.write( "Six years ago Sally's age = x-6 and Lisa's age = 1/9(x-6)\r
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\n" ); document.write( "\n" ); document.write( "Lisa's present age minus 6 years should = her age six years ago.
\n" ); document.write( "1/3x-6 = 1/9(x-6)
\n" ); document.write( "Remove the parenthesis.
\n" ); document.write( "1/3x-6 = 1/9x-6/9
\n" ); document.write( "Combining like terms.
\n" ); document.write( "1/3x-1/9x = 6-6/9
\n" ); document.write( "Using common denominators.
\n" ); document.write( "6/27x = 48/9
\n" ); document.write( "Divide both sides by 6/27.
\n" ); document.write( "x = 48/9 divided by 6/27 = 24\r
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\n" ); document.write( "\n" ); document.write( "Sally is 24 and Lisa is 1/3 of 24 or 8. Six years ago Sally was 18 and Lisa was 2.\r
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\n" ); document.write( "\n" ); document.write( "A tough problem?
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