document.write( "Question 550167: Several of us at work have attempted this with no success!
\n" ); document.write( "A father and son are aged 71 and 34 respectively, at what age was the father 3 times the age of his son and at what age will the son be half his fathers age?
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Algebra.Com's Answer #358488 by TutorDelphia(193)\"\" \"About 
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Given in the problem:
\n" ); document.write( "f=71
\n" ); document.write( "s=34
\n" ); document.write( "Let's have x be the number of years ago. So if it was x=10 years ago the father would be 61 and the son 24\r
\n" ); document.write( "\n" ); document.write( "3(s-x)=(f-x)
\n" ); document.write( "Plug in the father and son's ages
\n" ); document.write( "3(34-x)=(71-x)
\n" ); document.write( "Distribute the 3
\n" ); document.write( "102-3x=71-x
\n" ); document.write( "+3x to both sides
\n" ); document.write( "102=71+2x
\n" ); document.write( "-71 to both sides
\n" ); document.write( "31=2x
\n" ); document.write( "x=15.5
\n" ); document.write( "15 and a half years ago the son was 18.5 and the father was 55.5 . 55.5 is 3 times 18.5. The fact that it doesn't turn out evenly makes the problem a bit harder to figure out.\r
\n" ); document.write( "\n" ); document.write( "For the other half we start with a similar equation. However, I'm going to use y and have it represent years from now because this hasn't happened yet.
\n" ); document.write( "(1/2)(71+y)=(34+y)
\n" ); document.write( "Multiply everything by 2 to clear the fractions
\n" ); document.write( "71+y=68+2y
\n" ); document.write( "71=68=y
\n" ); document.write( "3=y\r
\n" ); document.write( "\n" ); document.write( "So in 3 years the son will be 37 and dad will be 74.\r
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