document.write( "Question 679031: (c) A is eleven times as old as B. In a certain number of years A will be five times as old as B, and five year after that A will be three times as old as B. How old are they now? \n" ); document.write( "
Algebra.Com's Answer #421780 by pmatei(79)\"\" \"About 
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\"A+=+11B\"
\n" ); document.write( "\"A+%2B+n+=+5%28B%2Bn%29\"
\n" ); document.write( "\"A+%2B+n+%2B+5+=+3%28B+%2B+n+%2B+5%29\"\r
\n" ); document.write( "\n" ); document.write( "In the last 2 equations replace A with 11B (first equation) and distribute on the right side of the equations:\r
\n" ); document.write( "\n" ); document.write( "\"11B+%2B+n+=+5B+%2B+5n\"
\n" ); document.write( "\"11B+%2B+n+%2B+5+=+3B+%2B+3n+%2B15\"\r
\n" ); document.write( "\n" ); document.write( "Bring the right side B-terms in the left side and the n's and 5 from the left side to the right side:\r
\n" ); document.write( "\n" ); document.write( "\"6B+=+4n\"
\n" ); document.write( "\"8B+=+2n+%2B+10\"\r
\n" ); document.write( "\n" ); document.write( "Divide by 2 the first equation:\r
\n" ); document.write( "\n" ); document.write( "\"3B+=+2n\"
\n" ); document.write( "\"8B+=+2n+%2B+10\"\r
\n" ); document.write( "\n" ); document.write( "Subtract first equation from the second one:\r
\n" ); document.write( "\n" ); document.write( "\"5B+=+10\"\r
\n" ); document.write( "\n" ); document.write( "\"B+=+2\"\r
\n" ); document.write( "\n" ); document.write( "\"A+=+11B+=+22\"\r
\n" ); document.write( "\n" ); document.write( "In \"n=3\" years \"A=25\" and \"B=5\" (A five times older than B), and five years after that \"A+=+30\" and \"B=10\" (A three times older than B).
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