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September 4

It's homework, but I need help!

We're doing systems of linear equations, which is easy when it's just the numbers, but the word problems are hard for me sometimes.

     The tens digit of a two-digit number exceeds the units digit by 5. 
     If the digits are reversed, the number that results is three times
     the sum of the digits. Find the original number.

I made the tens digit a and the units digit b. I got a=b+5 and 10b+a=3(a+b). The first one I made a-b=5, but on the second one, I'm at a loss. It looks like I shoulkd distribute the 3(a+b), wjich gives me 10b+a=3a+3b and you would subtract both 3a and 3b from both sides. But that leaves you with nothing on the right side. Unless I misunderstood the problem, I don't understand where I went wrong. Please help. schyler 14:28, 4 September 2006 (UTC)[reply]

You are doing it right:

a=b+5 (correct)

and From your second equation.

7b =2a

Subst for 'a' in this second equn.Hey presto!--Light current 14:40, 4 September 2006 (UTC)[reply]

But that's like b=(2/7)a and a=(2/7)a+5, which is really confusing, and doesn't look anything like we did in class. I need to put it in the form where a-b=5 and the second one where I can eliminate a top one, or at least where a and b are one side and a number on the other and I can do a matrix on my calculator. schyler 14:58, 4 September 2006 (UTC)[reply]

So, from a=(2/7)a+5, multiply everything by 7 and you have 7a=2a+35. Does that help ? Gandalf61 15:18, 4 September 2006 (UTC)[reply]

Ohhh. Okay. Thank you very much. I guess next time I shouldn't wait the entire weekend and then do it the day before school.schyler 15:37, 4 September 2006 (UTC)[reply]

No you shouldnt! 8-)--Light current 15:57, 4 September 2006 (UTC)[reply]

There is nothing wrong with having nothing on the right side:

10b+a = 3a+3b

Subtract right hand side from both:

(10b+a)−(3a+3b) = (3a+3b)-(3a+3b)

Simplify:

−2a+7b = 0

While I don't know what you learned in school, you can now proceed:

a−b = 5

Multiply both sides by 2:

2a−2b = 10

Add the other equation from above:

(2a−2b)+(−2a+7b) = 10+0

Simplify:

5b = 10

Divide both sides by 5:

b = 2

and now you can easily find the value of a. --Lambiam ^Talk 16:13, 4 September 2006 (UTC)[reply]

Erm, excuse me guys, but why do you keep a and b in your calculations? Utilise the first condition, and get rid of a immediately!

We have The tens digit ... exceeds the units digit by 5. — so a=b+5.

Then If the digits are reversed, the number that results... — substitiute a now and get: 10b+(b+5)

...is three times the sum of the digits. — ...= 3(b+(b+5)).

So we have an equation: 11b+5 = 6b+15, which gives the b value.

With this approach you can solve the "problem" in your mind, without writing a single digit.

Of course you might want to practice the formal approach — but in that case build the equation system:

a = b+5

10b+a = 3(a+b)

transform to some canonical form:

a-b=5

-2a+7b=0

and then solve it eg. with Gaussian elimination routine.

CiaPan 17:09, 4 September 2006 (UTC)[reply]

Because were trying to show the student how to do it his'/her way to avoid confusion. THere are many ways to solve this, but Gaussain elimination like using a hammer to crack an egg--Light current 20:42, 4 September 2006 (UTC)[reply]

Please note the implicit if in my previous entry: if you try to explain, teach, learn or practice the formal approach (which works also for systems of dozens or hundreds equations), then you should keep all steps necessary for complete solution. The Gaussian method (or some similar) certainly is not tailored for two-by-two problem, however it is a part of a general solution.

On the other hand if you try to explain or teach how to quickly solve simple systems, then point out how to simplify the problem. Reducing variables as soon as possible is more effective than transforming equations. Keeping 'the student's way' is not necessarily the best way, especially when you can teach the easier and faster one.
CiaPan 21:28, 4 September 2006 (UTC)[reply]

Yes but this person is obviously not at that level yet, so mentioning advanced solution methods is likely only to muddy the waters for him.her.--Light current 22:43, 4 September 2006 (UTC)[reply]

I'm sure this isn't how they want you to do it, but there are only 5 possibilities based on the first clue:

The tens digit of a two-digit number exceeds the units digit by 5.

This gives us: 94, 83, 72, 61, or 50. You could just try all 5 out with the remaining clue and avoid the use of any algebra. StuRat 04:25, 5 September 2006 (UTC)[reply]

Thays right- its not!. If you look at the question, (s)he states: we are doing systems of linear equations

so very few marks would be gained by the empirical approach. But I give you marks for proposing an alternative method in the freestyle category. 8-)--Light current 04:45, 5 September 2006 (UTC)[reply]

If we're in the business of solving the problem "freestyle", it's easier to ignore the first datum altogether, and use only the equation 2a = 7b; Since a ≠ 0 and a and b are digits, the only solution is a = 7, b = 2. -- Meni Rosenfeld (talk) 05:02, 5 September 2006 (UTC)[reply]

Yes, but how did you do that? Empirically in your head?--Light current 05:05, 5 September 2006 (UTC)[reply]

No, 2a is divisble by 7, so a is divisible by 7 (2 and 7 have no common factor), so a = 7. -- Meni Rosenfeld (talk) 05:27, 5 September 2006 (UTC)[reply]

Erm... You've skipped some options. Use of the Fundamental theorem of arithmetic is a good idea, but your application is incomplete.

7\mid a\;\land \;0\leq a\leq 9\ \Rightarrow \ a\in \{0,7\}

. So either {a,b} = {0,0} or {a,b}={7,2}. The first condition is still required to eliminate {0,0}. -- Fuzzyeric 03:13, 6 September 2006 (UTC)[reply]

My point was that somebody with no background in algebra or solving systems of linear equations can solve this problem, and I showed how. I happen to believe that the simplest solution method is best, with the obvious exception being when the teacher has asked you to use a certain method. Even then, I would still use the simple method as a check. StuRat 04:19, 6 September 2006 (UTC)[reply]

I've mentioned the fact that a ≠ 0. You only need to know that this is a two-digit number (the zeroth condition), not that a = b + 5.. -- Meni Rosenfeld (talk) 12:33, 6 September 2006 (UTC)[reply]

Isaac Newton and the Binomial theorem

The Newton article states: "In 1665 he discovered the generalised binomial theorem..." Westfall's Never at Rest describes the work which Newton began in the winter of 1664-5:

With the pattern now clear, he filled in all the intercalated columns of his initial table, corresponding to the fractional powers 1/2, 3/2, 5/2,... of the binomial (1 - x²). Could a more Wallisian procedure be imagined? Perhaps he might wish to check it empirically by computing the value of π that his series would yield? Not at all! He was perfectly confident. Calculating the well-known value of π was no challenge. It did not seriously occur to him that the principle of continuity on which he relied might betray him. Later he would recognize how flimsy the foundation was, and he would place the binomial expansion on a firmer footing...

Is this enough to date the discovery of the theorem? In October 1666 he would add the explanation:

This operation may bee continued at pleasure, y^e father the better. & from each terme ariseing from this operation may bee deduced a parte of y^e valor of y.

Is this quote the reason Westfall calls it a flimsy foundation? Should the article maybe have a weaker statement than "discovered the generalised binomial theorem"?EricR 17:07, 4 September 2006 (UTC)[reply]

I haven't read Westfall, nor the following article, but perhaps it will answer your questions: (D. T. Whiteside, "Newton 's Discovery of the General Binomial Theorem," Mathematical Gazette, 45 (1961), 175–80). --Lambiam ^Talk 06:29, 5 September 2006 (UTC)[reply]