document.write( "Question 550167: Several of us at work have attempted this with no success!
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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? \n" );
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Algebra.Com's Answer #358488 by TutorDelphia(193)![]() ![]() You can put this solution on YOUR website! 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 \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |