document.write( "Question 624211: A father is now 3 times older than his son.12 years ago he was 6 times as old as his son.Find their present Ages \n" ); document.write( "

Algebra.Com's Answer #392585 by josmiceli(19441) $\"\"$ $\"About$

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Let $\"+f+\"$ = father's age now
\n" ); document.write( "Let $\"+s+\"$ = Son's age now
\n" ); document.write( " $\"+f+-+12+\"$ = Father's age 12 years ago
\n" ); document.write( " $\"+s+-+12+\"$ = Son's age 12 years ago
\n" ); document.write( "given:
\n" ); document.write( "(1) $\"+f+=+3s+\"$
\n" ); document.write( "(2) $\"+f+-+12+=+6%2A%28+s+-+12+%29+\"$
\n" ); document.write( "----------------------
\n" ); document.write( "Substitute (1) into (2)
\n" ); document.write( "(2) $\"+3s+-+12+=+6s+-+72+\"$
\n" ); document.write( "Subtract $\"+3s+\"$ from both sides
\n" ); document.write( "(2) $\"+-12+=+3s+-+72+\"$
\n" ); document.write( "(2) $\"+3s+=+72+-+12+\"$
\n" ); document.write( "(2) $\"+3s+=+60+\"$
\n" ); document.write( "(2) $\"+s+=+20+\"$
\n" ); document.write( "and
\n" ); document.write( "(1) $\"+f+=+3s+\"$
\n" ); document.write( "(1) $\"+f+=+3%2A20+\"$
\n" ); document.write( "(1) $\"+f+=+60+\"$
\n" ); document.write( "The Father is 60 and the Son is 20
\n" ); document.write( "check:
\n" ); document.write( "(2) $\"+f+-+12+=+6%2A%28+s+-+12+%29+\"$
\n" ); document.write( "(2) $\"+60+-+12+=+6%2A%28+20+-+12+%29+\"$
\n" ); document.write( "(2) $\"+48+=+6%2A8+\"$
\n" ); document.write( "(2) $\"+48+=+48+\"$
\n" ); document.write( "OK \n" ); document.write( "

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