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.
Answer by Leaf W.(135)   (Show Source):
You can put this solution on YOUR website! B = B's age now
2B = A's age now ("A is at present twice the age of B")
B - 8 = B's age eight years ago
2B - 8 = A's age eight years ago
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"Eight years ago A was three times the age of B"
"'A' eight years ago = 3('B' eight years ago)"
2B - 8 = 3(B - 8)
Distribute the 3 into B and 8: 2B - 8 = 3B - 24
Add 8 to both sides: 2B = 3B - 16
Subtract 3B from both sides: -B = -16
Divide both sides by -1: B = 16 <== B's age now
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If A = 2B and B = 16, then A = 2(16)
A = 32 <== A's age now
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Therefore, at present, A is 32 and B is 16.
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