document.write( "Question 444766: A is at present twice the age of B. Eight years ago A was 3 times the age of B. Calculate their present ages. \n" ); document.write( "
Algebra.Com's Answer #306557 by Leaf W.(135)\"\" \"About 
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B = B's age now
\n" ); document.write( "2B = A's age now (\"A is at present twice the age of B\")
\n" ); document.write( "B - 8 = B's age eight years ago
\n" ); document.write( "2B - 8 = A's age eight years ago
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\n" ); document.write( "\"Eight years ago A was three times the age of B\"
\n" ); document.write( "\"'A' eight years ago = 3('B' eight years ago)\"
\n" ); document.write( "2B - 8 = 3(B - 8)
\n" ); document.write( "Distribute the 3 into B and 8: 2B - 8 = 3B - 24
\n" ); document.write( "Add 8 to both sides: 2B = 3B - 16
\n" ); document.write( "Subtract 3B from both sides: -B = -16
\n" ); document.write( "Divide both sides by -1: B = 16 <== B's age now
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\n" ); document.write( "If A = 2B and B = 16, then A = 2(16)
\n" ); document.write( "A = 32 <== A's age now
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\n" ); document.write( "Therefore, at present, A is 32 and B is 16.
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