document.write( "Question 402864: 18 yrs ago, a father was 3 times as old as his son. now the fathr is only twice as old as his son. then the sum of the present ages of the son and the father \n" ); document.write( "

Algebra.Com's Answer #284978 by josmiceli(19441) $\"\"$ $\"About$

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$\"f\"$ = Father's age now
\n" ); document.write( " $\"s\"$ = Son's age now
\n" ); document.write( "given:
\n" ); document.write( " $\"f+-+18+=+3%2A%28s+-+18%29\"$
\n" ); document.write( " $\"f+-+18+=+3s+-+54\"$
\n" ); document.write( " $\"3s+-+f+=+36\"$
\n" ); document.write( "and
\n" ); document.write( " $\"f+=+2s\"$
\n" ); document.write( "----------
\n" ); document.write( "By substitution:
\n" ); document.write( " $\"3s+-+2s+=+36\"$
\n" ); document.write( " $\"s+=+36\"$
\n" ); document.write( "and
\n" ); document.write( " $\"f+=+2s\"$
\n" ); document.write( " $\"f+=+72\"$
\n" ); document.write( " $\"36+%2B+72+=+108\"$
\n" ); document.write( "The sum of their ages now is 108
\n" ); document.write( "check answer:
\n" ); document.write( " $\"f+-+18+=+3%2A%28s+-+18%29\"$
\n" ); document.write( " $\"72+-+18+=+3%2A%2836+-+18%29\"$
\n" ); document.write( " $\"54+=+3%2A18\"$
\n" ); document.write( " $\"54+=+54\"$ \n" ); document.write( "

\n" );