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

if a>d>b was 1000000002>1000000001>1000000000, then wouldn't 1000000000/3000000003=1/3?

1/3 = 0.(3) but 1000000000/3000000003 is something like 0.(33333333000000000) which is smaller than 1/3

Main content

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?
(19 votes)
Answer
- Cary Wang
  Posted 9 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.
  (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?
(5 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.
  (6 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.
(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?
  (5 votes)
Ahmed Nazeer
Posted 9 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
(2 votes)
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.
  (4 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)
(3 votes)
Answer
Mohamed Ibrahim
Posted 3 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 ?
(1 vote)
Answer
- loumast17
  Posted 3 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
  (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.
(1 vote)
Answer
Ben Miller
Posted 6 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.
(1 vote)
Answer
- InnocentRealist
  Posted 6 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.
  (1 vote)
Agus
Posted 3 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
(1 vote)
Answer
Bella
Posted 10 years ago. Direct link to Bella's post “if a>d>b was 1000000002>1...”
if a>d>b was 1000000002>1000000001>1000000000, then wouldn't 1000000000/3000000003=1/3?
(0 votes)
Answer
- Piotr
  Posted 10 years ago. Direct link to Piotr's post “1/3 = 0.(3) but 100000000...”
  1/3 = 0.(3)
  but
  1000000000/3000000003 is something like 0.(33333333000000000) which is smaller than 1/3
  (6 votes)

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.

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