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Thread ID: 25163	2002-09-27 12:08:00	Elementary algebra	Deebee (1184)	Press F1

Post ID	Timestamp	Content	User
83802	2002-09-27 12:08:00	A "Peanuts" cartoon in the Dominion Post on 12 September convinced me I need to brush up on elementary algebra -- factoring etc. I was last taught it in 1964. Did some Googling on "algebra - lessons - tuition" and came up with hundreds of universities, colleges etc. offering courses. No free help. Any suggestions as to what exactly to search for? This was the problem: "A man has a daughter and a son. The son is three years older than the daughter. In one year the man will be six times as old as the daughter is now, and in ten years he will be fourteen years older than the combined age of his children. What is the man's present age?" I needed help to work out the answer.	Deebee (1184)
83803	2002-09-27 12:32:00	Well Deebee, that took me less than 30 seconds using Google, to get the answer 41	godfather (25)
83804	2002-09-27 12:52:00	Well I have the answer thanks to google but I'm buggered if I know how to work it out properly and I'm still at school. Normally I think I would of just used the guess and check method, but then again I am known for choosing the most hardest and awkward ways of doing things.	Sam H (525)
83805	2002-09-27 15:41:00	Not algebra - calculus, specifically quadratics let man's age = m and son's age = s and daughter = d then s = d +3 (son is three years older than daughter) m + 1 = 6d (man's age plus one equals daughter's age times six) m + 10 = (s +10) + (d +10) +14 (in ten years time, man's future age will be equal to son's future age plus daughter's future age plus fourteen) therefore, m + 10 = (s +10) + (d +10) +14 m + 10 = s + d + 34 m = s + d + 24 since s = d + 3, m = (d +3) + d + 24 m = 2d +27 multiply this by 3 3m = 6d + 81 now subtract m + 1 = 6d 3m = 6d + 81 m + 1 = 6d 2m - 1 = + 81 add one to each side 2m = 82 divide by 2 man's age = 41 Children's ages m + 1 = 6d 41 + 1 = 6d daughter's current age = 7 s = d +3 s = 7 + 3 son's current age = 10 To check m + 10 = (s +10) + (d +10) +14 41 + 10 = (10 + 10) + (7 +10) + 14 51 = 20 + 17 + 14 51 = 51	Merlin (503)
83806	2002-09-27 18:41:00	I disagree with the replied answer. however the maths could be correct. > This was the problem: > "A man has a daughter and a son. The son is three > years older than the daughter. In one year the man > will be six times as old as the daughter is now, and > in ten years he will be fourteen years older than the > combined age of his children. What is the man's > present age?" > "A man has a daughter and a son. The son is three > years older than the daughter. === > yes so far so good, (fact #2) In one year the man > will be six times as old as the daughter is now, === > yes could be true. and > in ten years he will be fourteen years older than the > combined age of his children. === > WHAT YEAR 2012 ??? What is the man's > present age?" === > in year 2002 OK WHAT IS THE RELATION YEAR DIFFERENCE BETWEEN THIS YEAR 2002 AND YEAR FROM FACT #2?	E.ric (351)
83807	2002-09-27 18:50:00	Oh silly me!! > (fact #2) In one year the man > will be six times as old as the daughter is now, or another way of putting it > (fact #2) NEXT year the man > will be six times as old as the daughter is now,	E.ric (351)
83808	2002-09-27 22:05:00	Very very good Merlin, you certainly are a wizard . AFAIK they do not teach maths like that in schools today . Well, certainly not up to the fourth form . I learnt stuff like that in form one/two but unfortunately, from lack of use, I've almost forgotten how to apply it . It's coming back to me now though! :D	Susan B (19)
83809	2002-09-27 22:26:00	> AFAIK they do not teach maths like that in schools > today . Well, certainly not up to the fourth form . bollocks . They do too . You start learning it in 3rd form, although in a somewhat milder form, and continue up til the 7th form . There are 3x3 simultaneous equations in 6th form maths, and again in 7th form stats AND in calc . I'm 2nd year uni now, and we learnt it AGAIN in year 1 intro calculus . . . . Well, there was when I was there anyway . . . . They TEACH it . That's not to say that anyone LEARNS it .	loser (538)
83810	2002-09-27 22:30:00	try searching for "tutorial" rather than "tuition" maybe add "free online" to your search too....	loser (538)
83811	2002-09-27 22:33:00	Sad they have to teach it through high-school... Like Susan B, I learned most of it in F1 and F2. Mike.	Mike (15)
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