document.write( "Question 907097: a boy's age is one third of his father's age. In ten years the son's age will be three less than half the father's age. Find the age of each.
\n" ); document.write( "B= boys age
\n" ); document.write( "F= fathers age
\n" ); document.write( "I found the variables to equal:
\n" ); document.write( "3b= f
\n" ); document.write( "B=1/3F \r
\n" ); document.write( "\n" ); document.write( "My equation was
\n" ); document.write( "10+b=1/2f-3
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Algebra.Com's Answer #550161 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
Let \"+f+\" = father's age now
\n" ); document.write( "Let \"+s+\" = son's age now
\n" ); document.write( "\"+f+%2B+10+\" = father's age in 10 yrs
\n" ); document.write( "\"+s+%2B+10+\" = son's age in 10 yrs
\n" ); document.write( "--------------------------------
\n" ); document.write( "(1) \"+s+=+%281%2F3%29%2Af+\"
\n" ); document.write( "(2) \"+s+%2B+10+=+%281%2F2%29%2A%28+f+%2B+10+%29+-+3+\"
\n" ); document.write( "---------------------------------
\n" ); document.write( "(2) \"+2%2A%28+s+%2B+10+%29+=+f+%2B+10+-+6+\"
\n" ); document.write( "(2) \"+2s+%2B+20+=+f+%2B+4+\"
\n" ); document.write( "(2) \"+f+-+2s+=+16+\"
\n" ); document.write( "and
\n" ); document.write( "(1) \"+f+=+3s+\"
\n" ); document.write( "Substitute (1) into (2)
\n" ); document.write( "(2) \"+3s+-+2s+=+16+\"
\n" ); document.write( "(2) \"+s+=+16+\"
\n" ); document.write( "and
\n" ); document.write( "(1) \"+f+=+3%2A16+\"
\n" ); document.write( "(1) \"+f+=+48+\"
\n" ); document.write( "-----------------
\n" ); document.write( "The father is 48
\n" ); document.write( "The son is 16
\n" ); document.write( "---------------
\n" ); document.write( "check:
\n" ); document.write( "(2) \"+s+%2B+10+=+%281%2F2%29%2A%28+f+%2B+10+%29+-+3+\"
\n" ); document.write( "(2) \"+16+%2B+10+=+%281%2F2%29%2A%28+48+%2B+10+%29+-+3+\"
\n" ); document.write( "(2) \"+26+=+%281%2F2%29%2A58+-+3+\"
\n" ); document.write( "(2) \"+26+=+29+-+3+\"
\n" ); document.write( "(2) \"+26+=+26+\"
\n" ); document.write( "OK
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