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### Course: Integrated math 3 > Unit 13

Lesson 5: Modeling with rational functions- Analyzing structure word problem: pet store (1 of 2)
- Analyzing structure word problem: pet store (2 of 2)
- Rational equations word problem: combined rates
- Rational equations word problem: combined rates (example 2)
- Rational equations word problem: eliminating solutions
- Reasoning about unknown variables
- Reasoning about unknown variables: divisibility
- Structure in rational expression

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# Reasoning about unknown variables

Sal solves the following challenge: Given that a>0, b<0, and a/b>a*b, what further can we learn about the variables? Created by Sal Khan.

## Want to join the conversation?

- at4:04why is b squared(4 votes)
- Jonathan,

You had

a/b < ab You multiplied each side by b to get

a*b/b < ab*b and the b/b is 1 and disappears so you get

a < ab * b which is

a < a *b * b and b*b is b² so

a < ab²

That is where the b² came from.

Then when you divide both sides by a you get

1 < b²(14 votes)

- When you divide A out at4:20do you always replace the empty side with 1?(4 votes)
- When you divide any variable out of a equation, you always replace the number or variable with 1. 1 is like the benchmark in math, it always takes the place of the number if the initial number is divided out.(4 votes)

- At4:27, can you take the square root of both sides? I tried that and it gives tells me that 1<b and -1<b. 1<b cannot be possible due to the given constraints. That leaves us with -1<b which is not what Sal concluded. What did I do wrong?(2 votes)
- Yes, we can take the square root of both sides, we just have to be very careful.

First of all, we know that the square root function is strictly growing over the interval [0, ∞).

This means that since 𝑏² > 1, then √(𝑏²) > √1

And because √1 = 1, we can write √(𝑏²) > 1

√(𝑏²) is the positive number whose square is equal to 𝑏².

Since 𝑏 is negative then −𝑏 is positive, and (−𝑏)² = 𝑏².

Thus, √(𝑏²) = −𝑏, which gives us the inequality

−𝑏 > 1

Multiplying both sides by −1 we get

𝑏 < −1(6 votes)

- I completely follow the tutorials, they make perfect sense when I watch him manipulate the expressions, but as soon as I try to do the practice problems my brain shuts off.(4 votes)
- Shouldn't we constrain a and b from the beginning by making them not equal to either 1, or -1? Because otherwise the first inequality is not true.

Ex:(a = 1, b = -1) ⇒ (1/-1) > (1*-1) ⇒ -1 > -1 : which is a false statement.(3 votes) - At3:52Sal changes the sign of the inequality saying that this is necessary, since we multiply both sides by a negative. Could anyone explain to me why this is the case? Like why do you need to change the sign and how do we know it, except from learning this as a rule?(2 votes)
- This is a often asked question in multiple videos.

So first way to think about it is finding values. If -x ≥ 3, put in values. If x=3, -3≥3 is false as would be any positive number. Try 0, 0≥3 still false. Try -3 and get -(-3)≥ 3 so 3≥3, finally true. The more negative we go, it would always stay true, so x ≤ -3.

Second way to show it, if -x ≥ 3, add x to both sides and subtract 3 on both sides to get -3 ≥ x, then flip the equation around to get x ≤ -3.(2 votes)

- Wouldn't the absolute value of 'b' also have to be greater than 'a' as the fraction -a/b is less than the multiplication product -(a*b)?(2 votes)
- No, you don't need the constraint that | b | > a.

Consider these examples.

a = 5 and b = -3. So "a" is larger than |-3|

5/(-3) is > 5(-3)

The key here is that dividing 2 integers will create a smaller negative number than multiplying them.

Hope this helps.(2 votes)

- at5:37how can b>1 when at the top of the screen it says b<0(0 votes)
- It can't.

And in the 30 seconds after5:37, Sal explains that b>1 cannot be part of the solution because b<0 is a restriction.(7 votes)

- I think we can also say that the absolute value of B must be greater than the absolute value of A. Am I right?(2 votes)
- Since we know b is negative, why don't we reverse the inequality when multiplying a/b * b?(2 votes)
- We did... at least, looking back at my notes I know I did. So, go ahead :)(1 vote)

## Video transcript

You go for a job interview. And the first thing that your
interviewer says is, look, you have great work experience. You seem like a
nice young person. But what I really care about
is your logical reasoning capabilities. So what she says
is, just sit down. I'm going to ask you a question
about some math expressions. And you say, sure, shoot away. Give me some questions
about math expressions. And she says, all right. So you have two integers,
integer a and integer b. And she tells you that
integer a is greater than 0, and integer b is less than 0. Then she says that
we also know that a/b is greater than a times b. So then she says, tell me
some interesting things about a and b and
a/b and a times b. And you say, OK, I'll
give my best shot. And I encourage you to
actually pause now and try to think about how these
things might relate, or how b might be
constrained, or a might be constrained,
and then unpause it. So I assume you've
unpaused it, so let's think about this a little bit. So the first thing, we know that
a is positive, b is negative. So if I take a positive
divided by a negative, what am I going to get? Well, this right over here, a
positive divided by a negative is going to be a negative. What happens if I take a
positive times a negative? Well, ab is also going
to be a negative. A positive times a negative
is also a negative. So really what we're
saying is that we have two negative quantities,
and this one right over here is greater than that one. So let's visualize
that on a number line. So let's say that this
right over here is 0. This right over here is
the positive direction. This is the negative direction. We know both of these are
negative, but a/b is larger. So a/b is going to be to
the right of a times b. So we know that a/b,
it has to be negative. It has to be the left of 0. a/b is going to be to
the right of a times b. So one way to think about it is
they're both negative, but a/b is going to be less negative. Or another way of
thinking about it, it's going to have a
smaller absolute value. Its distance from 0, which
is another way of thinking about absolute value, its
distance to the left of 0 is less than a times b's
distance to the left of 0. But because they're
both negative, and we're talking about
distances from the left of 0, the one that has a smaller
distance to the left of 0 is less negative, and is
therefore a greater number. So, so far your interviewer
seems impressed. So that was pretty good. You were able to get a
lot of clues just out of this little bit of
information that I gave you. But tell me more. Tell me more about
what b has to be and whether we can
constrain it in some way. And you say, well, OK. I've got some clues here, the
fact that this absolute value right over here is going to be
less than this absolute value. We know that the
absolute value of a/b is going to be less
than a times b, than the absolute
value of a times b. Once again, it's going
to be not as far left of 0 as a times b is. So that's how we can
make that statement. But let's actually manipulate,
algebraically manipulate this inequality here. We could multiply
both sides times b. So let's do that. Let's multiply
both sides times b. b Is less than the 0. So if you're
multiplying both sides of an inequality by
something less than 0, it swaps the inequality. That's why I got rid of it. I'm going to rewrite it. So b times a/b is
going to negative. We know it's negative. So it's going to be
less than ab times b. If we just multiply it
out, these cancel out. We get a is less than
a times b squared. Well, now if we
want to simplify it, we can divide both sides by a. And since a is
greater than 0, it does not change the inequality. So we divide both sides by a. This becomes 1, and this
just becomes b squared. So we're left with 1
is less than b squared. Or another way we could think
about this, we could say, well, if 1 is less than b squared,
that means that b-- let's be careful here-- that means
that the absolute value of b is going to be greater than 1. So this means that the absolute
value of b is greater than 1. And you say, wait,
Sal, how did you get from that to that
right over there? Well, just think about
this a little bit. If I square something, and
if it's greater than 1-- so that means that either
b is less than negative 1, because if it was negative
1, if you squared it, you would get 1. If it was greater than negative
1, or just directly greater than negative 1, if it was
negative 0.99, then when you square it, it would
be something less than 1. So that doesn't work. So b has to be less
than negative 1 or b has to be greater than 1, because
for the same exact logic. If it was exactly equal
to 1, if you squared it, this would be equal. If it was 0.5, then
squared it would be 0.25, which wouldn't
be greater than. So you know both of
these things are true. And this is another
way of saying that the absolute value
of b is greater than 1. Now, we have our
other constraint, that b is less than 0. So since we know
b is less than 0, we can take this
out of the picture. And if the absolute value
of b is greater than 1, and b is less than
0, then we know that b has to be
less than negative 1. So wherever negative 1 is, b
is going to be less than that. And your interviewer is very
impressed and she says, look, you've done very
good reasoning here. I think you deserve the job.