document.write( "Question 1771: Robert is 14 years old and his father is 38 years old. How many years ago was the father exactly seven time as old as his son? \n" ); document.write( "

Algebra.Com's Answer #581 by longjonsilver(2297) $\"\"$ $\"About$

You can put this solution on YOUR website!
this is a good question:-)\r
\n" ); document.write( "\n" ); document.write( "The one bit of information they want is \"when was the father x7 as old as his son.\" so, in algebraic terms, this is \"when is f=7r\" where f is the father and r is robert. Or, $\"%28f%2Fr%29=7\"$ \r
\n" ); document.write( "\n" ); document.write( "What we do know is that now, we have 38/14, which is NOT 7. Last year, we had 37/13, 2 years ago we had 36/12, 3 years ago we had 35/11 and \"n\" years ago we had $\"%2838-n%29%2F%2814-n%29\"$ and here we want this to be 7.\r
\n" ); document.write( "\n" ); document.write( "So, we just solve $\"%2838-n%29%2F%2814-n%29+=+7\"$ \r
\n" ); document.write( "\n" ); document.write( "You do the solving. Answer is n=10, ie ten years ago (when father was 28 and son was 4).\r
\n" ); document.write( "\n" ); document.write( "cheers
\n" ); document.write( "Jon.
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