document.write( "Question 173432: The four oldest people in the Golden City have lived a total of 384 years put together. The difference in ages for the youngest and second oldest is 14. The second youngest is 3 years older than the youngest. The oldest is 20 years older than the average of the second oldest and youngest. Find their ages and write from youngest to oldest. \n" ); document.write( "
Algebra.Com's Answer #128287 by Mathtut(3670)![]() ![]() ![]() You can put this solution on YOUR website! let the ages from youngest to oldest be: a, b ,c and d respectively \n" ); document.write( ": \n" ); document.write( "a+b+c+d=384.....eq 1 \n" ); document.write( "b=a+3...........eq 2 \n" ); document.write( "c-a=14..........eq 3--->c=a+14 revised eq 3 \n" ); document.write( "d=((a+c)/2)+20..eq 4 \n" ); document.write( ": \n" ); document.write( "take c's value from revise eq 3 and plug it into eq 4 so tht eq 4 is in terms of a. d=((a+(a+14))/2)+20--->d=(2a+14/2)+20...revised eq 4 \n" ); document.write( ": \n" ); document.write( "now take value of d from revised eq 4 and the value of b from eq 2 and the value of c in revised eq 3 and plug them into eq 1 \n" ); document.write( ": \n" ); document.write( "a+(a+3)+(a+14)+(2a+14)/2 +20=384: \n" ); document.write( ": \n" ); document.write( "3a+37+(2a+14)/2=384......multiply all terms by 2 to get rid of fraction. \n" ); document.write( ": \n" ); document.write( "6a+74+2a+14=768 \n" ); document.write( ": \n" ); document.write( "8a=680 \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "that some healthy living there...lol\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |