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 #30686 by longjonsilver(2297)\"\" \"About 
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OK, this is straight forward really. Just need to start correctly with a definition.\r
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\n" ); document.write( "\n" ); document.write( "Let x be the age of Lisa NOW.
\n" ); document.write( "Let y be the age of Sally NOW.
\n" ); document.write( "As Sally is 3 times older, then --> y = 3x\r
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\n" ); document.write( "\n" ); document.write( "6 years ago, the ages of the 2 were:
\n" ); document.write( "Lisa: x-6 and
\n" ); document.write( "Sally: y-6\r
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\n" ); document.write( "\n" ); document.write( "And Sally's age then was 9 times larger than Lisa's. So
\n" ); document.write( "(y-6) = 9(x-6)
\n" ); document.write( "y-6 = 9x-54
\n" ); document.write( "y = 9x-48\r
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\n" ); document.write( "\n" ); document.write( "So we have 2 equations:
\n" ); document.write( "y = 3x --> eqn 1 and
\n" ); document.write( "y = 9x-48 --> eqn2\r
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\n" ); document.write( "\n" ); document.write( "Now we have 2 equations to solve. There are a few ways of solving them. This is one:\r
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\n" ); document.write( "\n" ); document.write( "sub eqn1 into eqn2:
\n" ); document.write( "3x = 9x-48
\n" ); document.write( "0 = 6x-48
\n" ); document.write( "48 = 6x
\n" ); document.write( "x = 48/6
\n" ); document.write( "x = 8\r
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\n" ); document.write( "\n" ); document.write( "So Lisa is 8 years old now and so Sally is 24 years old.
\n" ); document.write( "and 6 years ago, Lisa was 2 and Sally 18... a difference of 9 times.\r
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\n" ); document.write( "\n" ); document.write( "jon.
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