Application Problem

Smith54

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The sum of two numbers is ten. The product of two numbers is twenty-one. Find the two numbers.

Let n = first number and n+1 = 10
Let n(n+1) = the product of the other numbwe.
n+1-10=0
n-9=0
n=9
9(9+1)=0
90
 

tkhunny

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Who said the numbers differed by one (1)?

First, this is a horribly worded problem statement. I sincerely hope this is not the original form.

The sum of two numbers is 10. Okay, 8 + 2 = 10.
The product of two number is 21. Okay, 3*7 = 21
Does it say they are the SAME two numbers?
Find the two numbers. Which two?

Let's assume we seek two values.

A = One of the values
B = The other value.

The sum of these two values is 10

A + B = 10

The product of these two values is 21

A * B = 21

Find the two values.

Are we getting anywhere?
 

Smith54

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I am not in the part of the book where they are discussing two different variables as part of the same equation. Its just a poorly worded problem. If you don't have anymore suggestions, I will just skip it.
 

mmm4444bot

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Yes, the wording is sloppy.

You're looking for two numbers whose sum is 10 and whose product is 21.

There's no need for variables, if you've memorized your multiplication table.

Simply consider factors of 21, and choose the pair that add up to 10.

 

tkhunny

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Re:

mmm4444bot said:

There's no need for variables, if you've memorized your multiplication table.
I cannot agree with that. They represent even greater abstractions in your head. You may as well learn to write them down and make them more concrete.
 
J

JeffM

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I agree that as the problem is posed it is a bad problem, but not because of wording. I presume that the problem statement said "the two numbers," which entails by the rules of English grammar that the numbers involved in the sum and the product are the same. Even if the definite article was omitted, it will be read into the problem. It is a bad problem because (a) an answer is obvious on inspection (to anyone who knows his or her multiplication tables) and therefore will teach nothing, and (b) because apparently the subject of equations in two variables has not yet been introduced.

Nevertheless, help has been requested on how to solve the problem, and I propose to give it.

Given (A + B) = 10 and (A * B) = 21, what are A and B?

So A = (10 - B).
So 21 = (A * B) = [(10 - B) * B] = 10B - B^2.
So B^2 - 10B + 21 = 0.

Does that look familiar?
 

masters

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Smith54 said:
The sum of two numbers is ten. The product of two numbers is twenty-one. Find the two numbers.
Hi Smith54,

My 2 cents....very succinctly,

\(\displaystyle \text{Let } x = \text{ the first number}\)

\(\displaystyle \text{Let } 10 - x = \text{ the second number}\)

\(\displaystyle \text{Their product is 21. Therefore, }\)

\(\displaystyle x(10 - x) =21\)
 

Denis

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Well, if Smitty is "learning to handle 2 variables", then he must be expected to use 2 variables:
a + b = 10 : b = 10 - a [1]
ab = 21 [2]
Substitute [1] in [2]:
a(10 - a) = 21
Carry on, Smitty!
 

tkhunny

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masters said:
\(\displaystyle \text{Let } x = \text{ the first number}\)

\(\displaystyle \text{Let } 10 - x = \text{ the second number}\)
This is an idea that I just don't understand - never have. If abstraction is the problem, and it usually is for beginners, this clearly is a greater level of abstraction than simply using two variables. One must create "10-x" by some internal process far more abstract than simply defining the second variable.

Nothing against the method. Obviously it is equivalent. Obviously it works for many, as it continues to appear in algebra text books. I just don't get it. It shouldn't exist.

My heretical views. I welcome others'.

Okay, that's enought of that. Back to work.
 

Denis

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tkhunny said:
My heretical views.
"heresay" is sufficient cause for da Pope to excommunicate you;
thus condemnation to an eternal afterlife of walking on coals...
 
J

JeffM

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tkhunny said:
masters said:
\(\displaystyle \text{Let } x = \text{ the first number}\)

\(\displaystyle \text{Let } 10 - x = \text{ the second number}\)
This is an idea that I just don't understand - never have. If abstraction is the problem, and it usually is for beginners, this clearly is a greater level of abstraction than simply using two variables. One must create "10-x" by some internal process far more abstract than simply defining the second variable.

Nothing against the method. Obviously it is equivalent. Obviously it works for many, as it continues to appear in algebra text books. I just don't get it. It shouldn't exist.

My heretical views. I welcome others'.

Okay, that's enought of that. Back to work.
I am not sure that "far more abstract" is quite right. Rather the method is more implicit and so quicker. I suspect that the substitution step suggested by Denis occurs almost subconsciously once you have done it a few hundred times. It SEEMS obvious once it is familiar. Now who writes text books in mathematics? Why teachers of mathematics, who are intimately familiar with the technique. The result is that the texts are obscure to those unfamiliar with the technique. Consequently, a huge percentage of even the educated population "hates" math. So I agree with you and Denis that the more cumbersome but more explicit method of using two variables and then substituting is a better method of teaching.

I am very new to this site, but what I like about it is that so many of the replies try to get at the root of the student's problem despite the risk of being condemned to walking on hot coals in the afterlife.
 

tkhunny

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In this case, the student received what appeared to be useful replies in the first couple of posts. It seemed, while we were at it, that some discrete discussion might be appropopriate. Frankly, I think it's at least a little useful for students to see that there likely is NOT just one way to approach a problem and that perfectly capable mathematics educators don't always agree on just what might be the best way to go.

I do not understand this "I am not sure that "far more abstract" is quite right. Rather the method is more implicit ..."

More implicit is more abstract. You cannot have it both ways.

Anyway, we seem to be close to the same page on this one. I'm not the official spokesman, but Welcome. You have already made a useful contribution, here. :D
 

mmm4444bot

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Gosh. I thought the exercise was beginning algebra.

I mean, that type of exercise that's generally assigned in preparation for lessons on how to factor a trinomial such as:

x^2 + 10x + 21

Are schools now teaching students to factor such a polynomial by solving a system of equations? :?

 

Denis

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Smith54 said:
The sum of two numbers is ten. The product of two numbers is twenty-one. Find the two numbers.
Let n = first number and n+1 = 10
Let n(n+1) = the product of the other numbwe.
n+1-10=0
n-9=0
n=9
9(9+1)=0
90
Well gee whizz et al, after seeing what Smitty tried up there,
methinks he ain't ready for anything except the very basics...
 

Smith54

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You all are emotionally disturbed, but I will continue to work with you all.
 

mmm4444bot

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Actually, everybody is a fool.

The only difference between us and you is that we know it. 8-)



The sum of two numbers is 15 and their product is 56. What are the two numbers?

 

Denis

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Smith54 said:
You all are emotionally disturbed, but I will continue to work with you all.
You will call on us "emotionally disturbs" to solve math problems?
Ain't you afraid of getting "disturbed" solutions? :roll:
 
J

JeffM

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Smith54 said:
You all are emotionally disturbed, but I will continue to work with you all.
That is very gracious of you. I am touched.
 

Smith54

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Let n= first number
Let n + n-1 = 10
Let n = the first number
Let n(n-1) = 21

n + n-1 =10
2n -1 = 10
2n-11 = 0
2n = 11
n = 11/2

n(n-1) =21
n^2 - n = 21
n^2 -n -21 =0

Not sure how to factor this. answer 3 and 7.
 

tkhunny

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Who said the numbers differed by one (1)?

RBGTHGANH

Round and round we go. If you want to get serious about it, please pay attention to what you are given. You can't just make it up. :?
 
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