When he works out b/c+d+b, couldn't he have plugged in numbers that fit the formula c>d>b? I used c=3, d=2, and b=1. b/c+d+b = 1/3+2+1 = 1/6 1/6<1/3 I have used this way before and it seems to work and be easier. Is his way a more full-proof way or is my way fine?

Well, this is an algebra problem. Sal expects you to solve it using algebraic expressions. That's the point of this video. Though your way of solving is very useful (I use it too), it is not an algebraic solution. So no, your way is not fine, at least in this case.

what does analytic mean?

When you analyse something. So you look at something and try to work out how it works.

If the values are negative then this ratio is larger than one third.

How can the number of animals be negative?

How did Sal know that d = 1/3 of the rectangle in his visual analysis? Couldn't d be a little less than 1/3 but greater than b? Could anyone explain? -- NAZ

`d` can be `1/3`. We don't know what it is exactly. All we really know is that `d` must be less than `1/2`. The biggest it could be is when `b` is zero. Then `d` could be almost `1/2` and `c` could be a tiny bit more than `1/2`.

the last reasoning that C+b is greater than 2b, it seems very reasonable and easy, but can we express that very simple reasoning by math or sometimes we have to write the logic in words ?

Do you mean x+d . 2b? Either way, it kind of already is an algebraic argument using inequalities. Basically we are given c>d>b from the start, so we can use that. so we can then use c+1 > b+1 and c+2>b+2 and keep goign for any number and eventually get c+b > b+b

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Algebra (all content)

Course: Algebra (all content) > Unit 13

Lesson 11: Modeling with rational functions

Analyzing structure word problem: pet store (2 of 2)

Name: Analyzing structure word problem: pet store (2 of 2)
Uploaded: 2013-06-26T22:17:55Z
Description: Sal solves a word problem about the unknown number of bears, cats, and dogs in a pet store. This is part 2 where Sal uses an algebraic reasoning.

CCSS.Math:

Sal solves a word problem about the unknown number of bears, cats, and dogs in a pet store. This is part 2 where Sal uses an algebraic reasoning. Created by Sal Khan.

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katherinekc10
Posted 10 years ago. Direct link to katherinekc10's post “When he works out b/c+d+b...”
When he works out b/c+d+b, couldn't he have plugged in numbers that fit the formula c>d>b?

I used c=3, d=2, and b=1.

b/c+d+b = 1/3+2+1 = 1/6

1/6<1/3

I have used this way before and it seems to work and be easier.
Is his way a more full-proof way or is my way fine?
Button navigates to signup pageComment on katherinekc10's post “When he works out b/c+d+b...”
(20 votes)
Answer
- Cary Wang
  Posted 10 years ago. Direct link to Cary Wang's post “Well, this is an algebra ...”
  Well, this is an algebra problem. Sal expects you to solve it using algebraic expressions. That's the point of this video. Though your way of solving is very useful (I use it too), it is not an algebraic solution. So no, your way is not fine, at least in this case.
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  (3 votes)
Rey #FilmmakerForLife #EstelioVeleth.
Posted 8 years ago. Direct link to Rey #FilmmakerForLife #EstelioVeleth.'s post “what does analytic mean?”
what does analytic mean?
Button navigates to signup pageComment on Rey #FilmmakerForLife #EstelioVeleth.'s post “what does analytic mean?”
(4 votes)
Answer
- Dan Reed
  Posted 8 years ago. Direct link to Dan Reed's post “When you analyse somethin...”
  When you analyse something.
  So you look at something and try to work out how it works.
  Comment on Dan Reed's post “When you analyse somethin...”
  (5 votes)
Ben Holt
Posted 8 years ago. Direct link to Ben Holt's post “If the values are negativ...”
If the values are negative then this ratio is larger than one third.
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(2 votes)
Answer
- H. H.
  Posted 7 years ago. Direct link to H. H.'s post “How can the number of ani...”
  How can the number of animals be negative?
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  (6 votes)
Tronax
Posted 6 years ago. Direct link to Tronax's post “i think there is a bit ea...”
i think there is a bit easier way to think about it, in general. ill show it here.

we are given that c > d > b. so, its clear that c > b and d > b
1. if c is greater than b, than we could say that c = b + k, where k is positive number
2. if d is greater than b, than we could say that d = b + j, where j is positive number

now we can substitude c and d: b/(c+b+d) as b/(b+k + b + b+j)
rearrange and sum the b's: b/(b+k + b + b+j) = b/(3b + k + j)

and now its simple, we know that b/3b > b/(3b + k + j) because adding positive numbers to denominator lowers the output.
b/3b = 1/3, so we get
1/3 > b/(3b + k + j)
substitude b and d back into denominator, and here is our solution:
1/3 > b/(c+b+d)
Button navigates to signup pageComment on Tronax's post “i think there is a bit ea...”
(3 votes)
Answer
- ɘɒbiᴎoƚʏT ᴎᴎʏlꟻ
  Posted 2 months ago. Direct link to ɘɒbiᴎoƚʏT ᴎᴎʏlꟻ's post “Yes, that is a valid appr...”
  Yes, that is a valid approach as well. You used the fact that c > b and d > b to express c and d in terms of b, and then substituted those expressions in the denominator. From there, you used the fact that adding positive numbers to the denominator of a fraction lowers its value, and compared the resulting fraction to 1/3. Finally, you substituted the expressions for c and d back into the denominator to get the final result. Well done!
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  (1 vote)
Ahmed Nazeer
Posted 10 years ago. Direct link to Ahmed Nazeer's post “How did Sal know that d =...”
How did Sal know that d = 1/3 of the rectangle in his visual analysis?
Couldn't d be a little less than 1/3 but greater than b?
Could anyone explain? -- NAZ
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(1 vote)
Answer
- Wrath Of Academy
  Posted 9 years ago. Direct link to Wrath Of Academy's post “`d` can be `1/3`. We don'...”
  d can be 1/3. We don't know what it is exactly. All we really know is that d must be less than 1/2. The biggest it could be is when b is zero. Then d could be almost 1/2 and c could be a tiny bit more than 1/2.
  Button navigates to signup page
  (3 votes)
Mohamed Ibrahim
Posted 4 years ago. Direct link to Mohamed Ibrahim's post “the last reasoning that C...”
the last reasoning that C+b is greater than 2b, it seems very reasonable and easy, but can we express that very simple reasoning by math or sometimes we have to write the logic in words ?
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(1 vote)
Answer
- loumast17
  Posted 4 years ago. Direct link to loumast17's post “Do you mean x+d . 2b? Ei...”
  Do you mean x+d . 2b? Either way, it kind of already is an algebraic argument using inequalities. Basically we are given c>d>b from the start, so we can use that. so we can then use c+1 > b+1 and c+2>b+2 and keep goign for any number and eventually get c+b > b+b
  Comment on loumast17's post “Do you mean x+d . 2b? Ei...”
  (2 votes)
Ahmed Nasret
Posted 4 years ago. Direct link to Ahmed Nasret's post “kindly judge this b/(b+c+...”
kindly judge this
b/(b+c+d) = 1/ ( (d+c)/b) +1 )
b,c and d should be integers thus min b is 1, min d is 2 and min c is 3
hence: the denominator expression ((b+c)/d)+1 min value is 6
the whole expression max value is 1/6, means the expression in subject must be lesser than or equal 1/6 which in all cases lesser than 1/3
whatever, Sal's way is better, because it covers the case of non-integers values of a,b and c. such proposal only allow us to put min values and tackle the way like i did.
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(1 vote)
Answer
- ɘɒbiᴎoƚʏT ᴎᴎʏlꟻ
  Posted 2 months ago. Direct link to ɘɒbiᴎoƚʏT ᴎᴎʏlꟻ's post “Your reasoning seems corr...”
  Your reasoning seems correct, but it is important to note that it only works for the specific values of b, c, and d that you provided. Sal's approach is more general and can be applied to any values of b, c, and d. Additionally, his approach uses algebraic manipulation, which is a more formal and rigorous method for comparing fractions with different denominators.
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  (1 vote)
Ben Miller
Posted 7 years ago. Direct link to Ben Miller's post “I am a little confused ab...”
I am a little confused about one of the practice problems. A>B>C, and it asks the relationship between B+C and 2A-B. The analytical solution says B+C<2A-B, but when I try it with real sets of numbers I get different results. Take {A,B,C} to be the sets {3,2,1} and {4,3,2} they produce different equality relationships.
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(1 vote)
Answer
- InnocentRealist
  Posted 7 years ago. Direct link to InnocentRealist's post “Maybe that's the problem ...”
  Maybe that's the problem where a>(b+c) is given in the problem.
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  (1 vote)
Agus
Posted 4 years ago. Direct link to Agus's post “i made it using infinity....”
i made it using infinity... much more complicated than Sal but here it goes.

"b tends to infinity,
as d>b, d tends to infinity+1,
and as c>d>b, c tends to infinity+2
so the denominator will have a bigger infinity than 3x(infinity)

thus b/(b+c+d) will always be <1/3."

is this reasoning correct? am i commiting conceptual mistakes?

thanks in advance
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(1 vote)
Answer
- ɘɒbiᴎoƚʏT ᴎᴎʏlꟻ
  Posted 2 months ago. Direct link to ɘɒbiᴎoƚʏT ᴎᴎʏlꟻ's post “Your reasoning is correct...”
  Your reasoning is correct in that the denominator, which tends to infinity + 3 as b tends to infinity, is much larger than 3 times infinity, so the fraction b/(b+c+d) tends to 0 as b tends to infinity. Therefore, b/(b+c+d) will always be less than 1/3.
  
  However, it's important to note that this reasoning relies on the assumption that c and d tend to infinity at a faster rate than b. While this may be true in this particular problem, it's not necessarily always true in other situations. So, it's important to be careful when using infinity in mathematical reasoning.
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  (1 vote)
nak
Posted 5 years ago. Direct link to nak's post “Here is my way of solving...”
Here is my way of solving this problem:
Take b/(c+d+b) divided by b and we have 1/[(c/b)+(d/b)+1]. Because c>b then c/b > 1. d>b then d/b > 1 too. 2 numbers bigger than 1 plus 1 will bigger than 3. And we have two rational numbers with the same numerator 1. The one have bigger denominator is smaller. 1/[(c/b)+(d/b)+1] < 1/3 => b/(c+d+b).
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(1 vote)
Answer
- ɘɒbiᴎoƚʏT ᴎᴎʏlꟻ
  Posted 2 months ago. Direct link to ɘɒbiᴎoƚʏT ᴎᴎʏlꟻ's post “Your approach is correct ...”
  Your approach is correct and similar to the approach taken by Sal in the original solution. By dividing both the numerator and denominator of the original expression by b, you obtained a new expression of the form 1/[(c/b)+(d/b)+1].
  
  Then, by observing that (c/b) and (d/b) are both greater than 1, you correctly noted that their sum plus 1 is greater than 3. Therefore, 1/[(c/b)+(d/b)+1] is less than 1/3.
  
  Finally, you concluded that b/(c+d+b) is greater than 1/3, which is the same as saying that it is less than or equal to 1/3 (since it is a positive fraction). Well done!
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  (1 vote)

Video transcript

In the last video, we made a visual argument as to why this expression has to be less than 1/3, and this expression we already figured out is the fraction that are bears. Now we will make an algebraic argument, or I could call it an analytic argument. And to make this argument, I'm going to leave this expression-- we know this is the fraction that are bears-- and I'm going to write this 1/3 in a form that looks a lot like this, and then based on the information we have, we can directly compare them. So how can I write 1/3? Maybe with the b as a numerator. Well, 1/3 is the same thing as b over 3b, which is the exact same thing as b over b plus b plus b. So now, this is looking pretty similar. The only difference between this expression right over here, b over c plus d plus b and b over b plus b plus b is that our denominators are different. And the only difference in our denominators, this denominator has a c plus d here, while this has a b plus b over here. Now, we have to ask ourselves a question. What is larger? Is c plus d larger than b plus b? And I encourage you to pause that and think about that for a second. Well, yes. We already see right over here. It was given to us that c is greater than d that is greater than b, so both c and d are greater than b. So c plus d is definitely going to be greater than b plus b. So this denominator right over here is greater, so this has a larger denominator. This right over here has a smaller denominator. And since we know this has a larger denominator, this has a smaller denominator, they have the exact same numerator-- they both have b as a numerator-- we know that this whole thing must be a smaller quantity. If you have the same numerator but one expression has a larger denominator, it must be smaller. Wait, so how does that work? Well, just remember. I mean, just imagine. You have the same numerator, what's going to be bigger, a over 7 or a over 5? Well here, you're dividing a by 7. You're dividing into many more chunks than over here, so this right over here is smaller. This right over here is larger. So this is the larger. This right over here is smaller. So the same numerator, the larger the denominator, the smaller the quantity is going to be. So going back to the original question, this is the smaller quantity, and this right over here, 1/3, is the larger quantity.