document.write( "Question 1059463: The sum of the ages of Mark's two sons is equal to his own age of 51.
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document.write( "21 years ago, Mark's older son was twice as old as the younger son.
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document.write( "How old was Mark when the sum of his sons' ages was half his age? \n" );
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Algebra.Com's Answer #674546 by addingup(3677)![]() ![]() You can put this solution on YOUR website! a+b = 51 so a = 51-b \n" ); document.write( "a-21 = 2(b-21) \n" ); document.write( "a-21 = 2b-21 substitute for a: \n" ); document.write( "51-b-21 = 2b-21 \n" ); document.write( "72 = 3b flip it \n" ); document.write( "3b = 72 \n" ); document.write( "b = 24 and a = 51-b = 51-24 = 27 \n" ); document.write( "------------------------------ \n" ); document.write( "check: 21 years ago: \n" ); document.write( "a = 27-21 = 6 \n" ); document.write( "b = 24-21 = 3 and 3 is 1/2 of 6 Correct answer. \n" ); document.write( "----------------------------- \n" ); document.write( "Mark is 51 and 51/2= 25.5. The boys are 3 years apart, as I just showed. Let's call the age of the youngest x, and the oldest will be x+3. So: \n" ); document.write( "x+x+3 = 25.5 \n" ); document.write( "2x = 22.5 \n" ); document.write( "x = 11.25 the youngest was 11 1/4 years old. \n" ); document.write( "------------------------------- \n" ); document.write( "The youngest is now 24 and 24-11 1/4 = 12 3/4 \n" ); document.write( "Mark is now 51 and 51-12 3/4 = 38 1/4 This is how old Mark was, 38.25 or 38 1/4 \n" ); document.write( ": \n" ); document.write( "John \n" ); document.write( " |