document.write( "Question 1065161: The second of three members are three times the first and the third is four more than one third the first. the sum of this members is 43. find the three numbers \n" ); document.write( "
Algebra.Com's Answer #680276 by JulietG(1812)\"\" \"About 
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Let the members be represented by A, B, and C.
\n" ); document.write( "What do we know from the question?
\n" ); document.write( "A + B + C = 43
\n" ); document.write( "B = 3A
\n" ); document.write( "C = .33A + 4
\n" ); document.write( "Substitute the values of the second two equations into the first.
\n" ); document.write( "A + B + C = 43
\n" ); document.write( "A + (3A) + (.33A+4) = 43
\n" ); document.write( "4A + .33A + 4 = 43
\n" ); document.write( "4.33A + 4 = 43
\n" ); document.write( "4.33A = 39
\n" ); document.write( "Divide.
\n" ); document.write( "A = 9 (disregard the insignificant places; due to rounding 1/3 to .33)
\n" ); document.write( "If we now know that A = 9, we can figure out the others.
\n" ); document.write( "B = 3A
\n" ); document.write( "B = 3(9)
\n" ); document.write( "B = 27
\n" ); document.write( "C = .33A + 4
\n" ); document.write( "C = 3 + 4
\n" ); document.write( "C = 7
\n" ); document.write( "Prove it by adding the values together.
\n" ); document.write( "9 + 27 + 7 = 43
\n" ); document.write( "Success! \n" ); document.write( "
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