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## Precalculus

### Unit 4: Lesson 5

Modeling with rational functions- Analyzing structure word problem: pet store (1 of 2)
- Analyzing structure word problem: pet store (2 of 2)
- Combining mixtures example
- Rational equations word problem: combined rates
- Rational equations word problem: combined rates (example 2)
- Mixtures and combined rates word problems
- Rational equations word problem: eliminating solutions
- Reasoning about unknown variables
- Reasoning about unknown variables: divisibility
- Structure in rational expression

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

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.

## Want to join the conversation?

- 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)- 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)

- When you analyse something.

So you look at something and try to work out how it works.(6 votes)

- If the values are negative then this ratio is larger than one third.(2 votes)
- How can the number of animals be negative?(5 votes)

- 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)`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)

- 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) - 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)
- 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)

- 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) - 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)
- Maybe that's the problem where a>(b+c) is given in the problem.(1 vote)

- 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) - if a>d>b was 1000000002>1000000001>1000000000, then wouldn't 1000000000/3000000003=1/3?(0 votes)
- 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.