Ben and Bill
Date: 12/21/97 at 10:21:53 From: SHAMALA Subject: Algebra Bill + Ben's Age = 91 Bill is twice as old as Ben was when Bill was as old as Ben is now. I've tried to form an equation but I get stuck in the first phrase. Please help. Thank you, Shamala
Date: 01/06/98 at 13:32:06 From: Doctor Dawn Subject: Re: Algebra Dear Shamala, Wow! This was a tough question and had me stumped for a while. At first I was trying to assign Ben a variable and Bill a different variable, but I couldn't figure out what to put down for the second sentence. Since the difference in their ages is constant, I finally decided to assign the difference in their ages a variable. I also know that Bill is older. Here goes: First, define the variables x = difference in Bill's and Ben's ages b = Ben's age This means b + x = Bill's age and Bill - x = Ben's age. (This will be important later) Next write the English sentences as math equations using the variables. Bill + Ben = 91 (b + x) + b = 91 so 2b + x = 91. (Equation 1) Now for the second sentence - Bill is twice as old as Ben was, when Bill was as old as Ben is now. The first Bill is now in the present and is represented by b + x The second Bill is in the past when he was Ben's age now which is b. Ben in the past was x years younger than Bill in the past, so Ben in the past is represented by: Bill (past) - x = b - x. The second equation then goes like this. Bill = twice as old as Ben was (b + x) = 2 * (b - x) (Equation 2) Now take both equations and solve. 2b + x = 91 b + x = 2*(b - x) Working with equation 2: b+x = 2b-2x (distribute the 2) 3x = b (add 2x and subtract b on each side) Substituting 3x for b into equation 1: 2*3x + x = 91 6x + x = 91 7x = 91 x = 13 The difference in their ages is 13. Using equation 1 again, and knowing x = 13: 2b + 13 = 91 2b = 78 b = 39 (Ben's age) Bill = b + x = 39 + 13 = 52. Bill is 52 years old; Ben is 39 years old. To check, when Bill was 39 (Ben's age now), Ben was 26. 52 is twice 26. -Doctor Dawn, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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