Rational Inequalities

Summary

This instructional video teaches viewers how to solve rational inequalities using the test point method. The presenter simplifies the process by treating inequalities as equations to identify critical values, then selects test points on a number line to determine intervals that satisfy the inequality. The video provides step-by-step solutions to various examples, emphasizing the importance of common denominators, critical values, and sign charts in finding the solution set for each inequality. The method is demonstrated with increasing complexity, offering a clear understanding of solving rational inequalities.

Takeaways

• 📝 Rational inequalities can be solved using the test point method.
• 🔍 Treat the inequality like an equation to find critical values.
• 📊 Use test points on a number line to determine which intervals satisfy the inequality.
• ✂️ Factor the denominator and numerator to identify excluded and critical values.
• ❌ Exclude values that make the denominator zero from the solution set.
• 🔢 Set up the number line by cutting it at critical values to test intervals.
• ✅ Identify intervals where the left-hand side of the inequality is positive or negative.
• 🔄 Use extreme values and simple test points to determine the sign of intervals.
• 🔧 For more complex inequalities, rearrange terms to get zero on one side.
• 📉 Combine fractions with a common denominator when necessary.
• 🔑 Use sign charts to visualize and verify solutions for rational inequalities.
• 🔢 Critical values are points where the numerator or denominator equals zero.
• 🧠 Plug in values within intervals to test whether the inequality holds true.
• 📈 The solution set includes intervals where the inequality is satisfied, using union for multiple intervals.
• 🧩 Simplify expressions and factor to make solving and testing easier.

Q & A

What is the main method discussed in the video for solving rational inequalities?

The main method discussed in the video for solving rational inequalities is the test point method.

What are critical values in the context of solving rational inequalities?

Critical values are the values that make the numerator or the denominator of a rational expression zero, as they are important in determining the intervals on the number line that satisfy the inequality.

Why is it necessary to treat the inequality like an equation when finding critical values?

Treating the inequality like an equation helps in finding the values that make the numerator or denominator zero, which are essential for identifying critical points and excluded values.

What is the first step in solving the given example inequality x - 2/x^2 - 36 > 0?

The first step is to factor the denominator x^2 - 36 into (x + 6)(x - 6) and identify the critical values where the denominator and numerator are zero.

What are the excluded values for the example inequality x - 2/x^2 - 36 > 0?

The excluded values for the example inequality are x = -6 and x = 6, as these values make the denominator zero.

How does the video suggest to determine the intervals that satisfy the inequality?

The video suggests using test points on a number line between the critical values to determine which intervals make the inequality true.

What is the purpose of using test points in solving rational inequalities?

Using test points helps to determine the sign of the rational expression in different intervals on the number line, which in turn helps to identify where the inequality holds true.

How does the video handle the case where the right-hand side of the inequality is not zero?

The video suggests adding or subtracting the non-zero value to both sides to create a zero on one side, which simplifies the process of identifying critical values and solving the inequality.

What is the solution set for the inequality x + 9/x + 7 ≤ -4 after adjusting the inequality to have zero on one side?

The solution set for the adjusted inequality is the interval from -37/5 to -7, including -37/5 but not including -7.

Why is it important to create a common denominator when combining fractions in an inequality?

Creating a common denominator is important because it allows for the simplification of the inequality into a single rational expression, making it easier to identify critical values and solve the inequality.

How does the video approach the inequality 4/x - (2 - 3)/x ≥ 0?

The video approaches this inequality by combining the fractions on the left-hand side over a common denominator and then using test points to determine where the inequality is satisfied.

What is the solution set for the inequality x > 4x/5 - x?

The solution set for this inequality is the union of the intervals 0 to 1 and 5 to infinity.

Why is factoring useful when solving rational inequalities?

Factoring is useful because it simplifies the process of identifying critical values and helps in creating a sign chart to determine where the inequality is true.

Outlines

00:00

📚 Introduction to Solving Rational Inequalities

In this video, the presenter introduces the concept of solving rational inequalities using a method called the Test Point method. The method involves treating the inequality as an equation to find critical values and then using test points on a number line to determine which intervals satisfy the inequality. The presenter chooses a simple inequality, x - 2/(x^2 - 36) > 0, to demonstrate the process. The critical values are identified as -6, 2, and 6, which are points where the expression is undefined or zero. The presenter then sets up a number line, marks these critical values, and uses test points to find the intervals where the left-hand side of the inequality is positive, which are the solutions to the inequality.

05:02

🔍 Analyzing Rational Inequalities with Variations

The presenter continues by tackling variations of rational inequalities. The second example involves an inequality with a non-zero right-hand side, x + 9/(x + 7) ≤ -4. To apply the Test Point method, the presenter manipulates the inequality to have zero on the right side by adding 4, resulting in a common denominator for combining the terms. The critical values are identified as -7 and -37/5, and the presenter uses test points to find the intervals where the left-hand side is less than or equal to zero. The solution set includes the interval from -37/5 to -7, excluding -7.

10:02

📈 Solving Rational Inequalities with Common Denominators

In the third example, the presenter solves an inequality with a zero right-hand side, 4/x - 2 - 3/x ≥ 0. The presenter combines the fractions with a common denominator, x(x - 2), resulting in a simplified inequality. The critical values are 0 and 2, and the presenter uses test points to identify where the left-hand side is greater than or equal to zero. The solution includes two intervals: -6 to 0, including -6 but not 0, and 2 to infinity, excluding 2.

15:03

🎯 Advanced Techniques for Rational Inequalities

The final example presented is an inequality involving a rational expression greater than a variable, x > 4x/5 - x. The presenter discusses different approaches, including subtracting x to the right side and factoring out a negative from the denominator. The presenter then combines the fractions with a common denominator and simplifies the inequality to find critical values at 0, 1, and 5. Test points are used to determine the intervals where the left-hand side is positive, resulting in a solution set of 0 to 1 and 5 to infinity. The presenter emphasizes the importance of common denominators, critical values, factoring, and sign charts in solving rational inequalities.

Keywords

💡Rational Inequality

A rational inequality involves a rational expression, which is a fraction where the numerator and/or the denominator are polynomials. The inequality indicates that the fraction is less than, greater than, less than or equal to, or greater than or equal to a certain value. In the video, the inequality x - 2 / (x^2 - 36) > 0 is an example.

💡Test Point Method

The test point method is a technique used to determine which intervals on a number line make a given inequality true. By selecting points within these intervals and substituting them into the inequality, one can check the sign of the expression. The video demonstrates this method by picking points like 0 and 1000 to test the inequality.

💡Critical Values

Critical values are specific points on the number line where the rational expression is either zero or undefined. These values divide the number line into intervals for testing. In the video, -6, 2, and 6 are identified as critical values for the inequality x - 2 / (x^2 - 36) > 0.

💡Number Line

A number line is a visual representation used to plot the critical values and intervals for testing the inequality. The video uses a number line to separate the regions and evaluate which intervals satisfy the inequality.

💡Undefined

An expression is undefined at points where the denominator equals zero, making the fraction invalid. In the video, x = -6 and x = 6 are points where the denominator x^2 - 36 becomes zero, making the expression undefined.

💡Numerator

The numerator is the top part of a fraction. When it equals zero, the entire fraction equals zero. In the video, the numerator x - 2 is zero when x = 2, which is one of the critical values.

💡Denominator

The denominator is the bottom part of a fraction. When it equals zero, the fraction is undefined. The video factors the denominator x^2 - 36 into (x + 6)(x - 6) to find the critical values where the expression is undefined.

💡Factor

Factoring is the process of breaking down an expression into a product of simpler expressions. In the video, the denominator x^2 - 36 is factored into (x + 6)(x - 6) to find the critical values.

💡Sign Chart

A sign chart is a tool used to determine the sign of an expression in different intervals on the number line. The video uses a sign chart to organize test points and determine where the inequality is true.

💡Union

Union is a mathematical operation that combines multiple sets. In the video, the solution to the inequality is given as the union of intervals (-6, 2) and (6, ∞), indicating all x-values that satisfy the inequality.

Highlights

Introduction to solving rational inequalities using the test point method.

Treating the inequality like an equation to find critical values.

Using test points on a number line to determine intervals that satisfy the inequality.

Solving the rational inequality x - 2 / x^2 - 36 > 0 as a starting example.

Factoring the denominator x^2 - 36 to find excluded values of -6 and 6.

Identifying critical values that make the numerator or denominator zero.

Setting up a number line with critical values at -6, 2, and 6.

Using test points to determine where the left-hand side is positive.

Finding that the left-hand side is positive between -6 and 2.

Using extreme values like -1000 and 1000 as test points.

Determining the solution set for the inequality x - 2 / x^2 - 36 > 0.

Modifying the inequality x + 9 / x + 7 ≤ -4 by adding 4 to both sides.

Combining fractions with a common denominator to simplify the inequality.

Identifying new critical values -7 and -37/5 and solving the modified inequality.

Using test points to find intervals where the left-hand side is less than or equal to zero.

Solving the inequality 4 / x - (2 - 3) / x ≥ 0 by combining fractions.

Creating a sign chart to determine where the left-hand side is greater than or equal to zero.

Finding the solution set for the inequality with intervals and critical values.

Approaching the inequality x > 4x / 5 - x by moving the entire fraction to the left.

Factoring the numerator and denominator to identify critical values of 0, 1, and 5.

Using test points to find intervals where the left-hand side is positive.

Final solution set for the inequality x > 4x / 5 - x is 0 to 1 and 5 to infinity.

Emphasizing the importance of common denominators, critical values, factoring, and sign charts in solving rational inequalities.

Transcripts

hey in this video we're going to solve

rational

inequalities you know less than or

greater than with rational expressions

or fractions and the way we're going to

do this is we're going to we're going to

use what's called a test Point method so

uh we're going to we're going to treat

the inequality like an equation to find

critical values and then we're going to

use test points on a number line between

those critical values to figure out

which intervals make the inequality true

let's get into

it let's solve the rational inequality x

- 2/ x^2 - 36 is greater

than

0 and I chose a simple one to get

started the reason I'm telling you it's

simple is because the right hand side is

already zero and the left hand side has

a single fraction those are those are

things we like

now my goal is to figure out when this

left hand side is greater than zero when

when is the left hand

side positive that's what I'm looking

for

okay because and I'm trying to figure

out the X values that make that true so

what I'm going to do is treat this like

an equation and ask myself okay I need I

need to figure out the excluded values

so you know if I if I think about the

left side I can factor that denominator

x^2 - 36

as x + 6 * x -

6 and I can see from this that -6 and

positive 6 will make the denominator

zero so those are excluded values

-6 and positive

6 I don't want X to equal

those and then my numerator to make it

zero is xals 2 so that's not I can I can

have zero on the top now I'm not saying

that's the solution to the inequality

I'm just saying these three values are

my critical

values values that make the top or the

bottom zero are critical values because

they are important numbers on the number

line when we start to figure out which

parts of the number line or which

intervals on the number line make the

INE equality true so let's set up our

number line

cut at those critical values and see

what we can find so um these are X

values I need to cut them in cut it in

order so -6 is the first of those three

values that occurs and then two and then

six and I already know um I need to

figure out what happens to the left hand

side for the X values I'm going to add

here so up above X I'll write left hand

side and kind of keep track of what

happens we already know know that when X

is -6 the left hand side this fraction

is undefined so I'm going to put a u

there if x is 2 the numerator is zero

and we know that when the numerator of a

fraction is zero the whole fraction is

zero when X is six the denominator is

zero so the left hand side is

undefined so what I've done is I've cut

my numberline these critical places

these key places and now I need to look

at all right what I happens to the left

hand side in between those critical

values so what I mentioned before is

using test points so I'm going to pick

an x value in each of these intervals to

plug in to the left hand side and see

what I get so for example between -6 and

two I I would plug in

zero and if I plug in zero the top is

going to be

negative 0 - 2 is

negative the bottom left 0 + 6 is

positive the bottom right 0 - 6 is

negative so the bottom is a positive

times a negative which is negative and a

negative divided by a negative is a

positive so this the left hand side is

positive when X is between -6 and

2 now what I'm interested in when is the

left hand side positive anyway so the

left hand side is positive here so I def

Ely I'm going to include all of those X

values between -6 and 2 in my solution

set okay now I'm going to move a little

quicker now you can use a calculator to

plug these values in or you can do it

mentally like I I've done it here um on

the the ends I always like to use

extreme values so I'm going to use

negative 1,000 and positive 1,000 for my

test

points and then between two and six

maybe I'll just use three

all right if I plug in negative

1000 we get a negative over a negative

times a

negative that all simplifies to be a

negative so I'm not going to shade this

interval because I want the intervals

that are greater than

zero if I plug in three I get a positive

over a positive time a negative so

that's

negative and if I plug in 1,00 I get a

positive over a positive time a POS

positive which is a positive and so I do

want to shade all the X values bigger

than six now we have to look at the

critical values

themselves -6 if I plug that in I get

undefined that is not positive so I

don't want to include that so I'll

barricade it with this open circle if x

is 2 I get equal to zero that is not

greater than zero so I don't want to

include two in my solution either and

six I get undefined I don't want to

include that so my solution the X values

that make this true are in the

intervals -6 to 2 or 6 to infinity and

since it's this interval or this

interval I put a union between those and

that is my solution to the inequality

these are all the X values that make

this inequality

true now that was a nice one again

because we had zero on one side and

single fraction on the left side let's

let's try some variations of

this okay let's look at x + 9 over x +

7 is less than or equal to

-4 now this one almost looks as nice as

the last one and you're just saying well

ne4 is not zero that's actually a big

problem because the technique we're

using has this check the

sign positive or negative to get our

solution set and that's why we need to

know where zero is important because

zero separates the positives from the

negatives so what we have to do is add

this four over to get a zero on the the

right

side and just like when I solve rational

equations I'm going to have to combine

these two guys with a common denominator

so I'm going to call four four over

one and figure out how to get a common

denominator now on the right hand side

I've still got less than or equal to and

then

zero so my common denominator between

the two fractions well I can multiply

the second fraction on top and bottom by

x +

7 and so that gives

me the same denominator of x + 7 when I

put these together and I get x + 4x

which is 5X I'm adding my numerators now

and I'm just Distributing this four in

my head and then I get 9 + 28 which is

uh

37 and now this is set up just like the

last example we've got zero on one side

in a single fraction on the left side

and our critical values

here well my denominator critical value

is well X can't be

-7 and my numerator if I get X by itself

I'd subtract

37 and divide by 5 I get X =

-37 FS which is like

-7.4 okay so now I know where to cut my

number

line my sign chart

here I need to cut it Well - 37 fths is

further left on my number line the

-7 and I need to see what happens to the

left hand side side I'm going to go

ahead and say when X is -7 well that

makes the fraction or the left hand side

undefined and when X is -37 fths it

makes the numerator zero which makes the

entire fraction zero the the entire left

side zero and then I'm going to use test

points again so I'll plug in negative

1000 so I get a negative on top that's

pretty easy see a negative on bottom

which gives me a

positive uh between these guys maybe

I'll plug in like

-7.1 which is going to give me a let's

see here a positive on top and a

negative on bottom so I get

negative and over here if I plug in a th

I get a positive on top the positive on

bottom which is

positive and I want to find where the

left hand side is less than or equal to

zero well it's equal to zero here it's

less than zero

this interval and it's not less than or

equal to Z at -7 because it's undefined

so my solution set is the interval from

- 37 fths to -7 including - 37 fths but

not

including

-7 so that is our solution for this

inequality just a slight upgrade and

complexity by having to get zero on one

side side and combining the fractions

with a common

denominator let's do another slightly

different example let's look at 4 overx

-

2 - 3

overx is greater than or equal to

zero and in this case I've got zero on

the right side that's good I just need

to combine my fractions with a common

denominator so I'll multiply top and

bottom of this fraction by X

and I'll mply top and bottom of this

fraction by x -

2 so when I put them together they're

all going to be over the common

denominator of x * x - 2 and I get 4X

here and remember this negative

distributes with the three so I get min

- 3x so 4x - 3x is just X and then

plus six out of

that and that's that's there is to it we

can see that our critical

values are 0er and two and we don't want

X to be either of those and my

numerator get xal

-6 we make a sign chart here to see what

happens to the left hand

side so let's see -6 is first and then

zero and then two

those are our X values let's see what

happens to the left hand side well if x

is -6 my numerator is zero so my

fraction is zero but if x is z or two my

denominator is zero so the left hand

side is

undefined and then we'll use some test

points let's start over here on the

right hand side by plugging in 1,00 for

X we get

positive over positive * positive so

that's

positive if we plug in one for X we get

a positive a positive * a negative

that's

negative if we plug in -3 for X we get

positive over negative * negative that's

positive and if we plug in 1000 for X we

get Negative over negative * negative

which is

negative and we're looking for where the

left hand side is greater than or equal

to

zero so it's equal to zero here it's

greater than zero there and it's greater

than zero there so let's

shade this critical value let's shade

this interval exclude this critical

value exclude this critical value and

shade this

interval so we got two intervals we

indicate that with the

Union -6 to 0 including -6 but not zero

Union 2 to Infinity not including either

one okay let's let's do one more example

just just to be thorough here I want you

to feel like you've got the idea down

really

well okay let's try this guy um X is

greater than the rational

expression 4X over 5 -

x and so we've got a couple different

ways we can do this we can subtract the

X over to the right

that's probably what I would do because

I'd rather the X be the negative then

rather make this whole fraction negative

um but let's go ahead and see what

happens if we move the entire fraction

over to the

left I need to I need to make X into a

fraction so I can combine

them and I'm just going to mention some

things here as a as a mathematician and

a

teacher we

typically we typically don't like

expressions like this I mean there's

nothing wrong with them um but it's it's

natural for uh a mathematician let me

see if I can get rid of that there we go

to think of that just off to the side

it's the same thing as Negative X +

5 and because there's a negative in the

front you know lots of times it's useful

to factor that negative

out of both terms which changes their

signs

and in a fraction that's nice because we

don't like to have a negative on the

bottom so what I'm going to

do you don't have to do this I just want

to I'm teaching math so I'm I want to

kind of tell you how MA a mathematician

might think about this I'm G to I'm

going to go ahead and write my

denominator as negative times the

quantity xus 5 I've just factored the

negative

out and in a fraction when you got a

Nega

Factor you can just pull it out of the

fraction because I don't want to get

into it too much but if you got a

negative on the top or the bottom you

can just pull it out to the front and so

if you do that minus a negative turns

into plus a positive so now we have X

over 1 + 4x over x -

5 you don't that none of that's

necessary but it's it's very common to

do that and it it just makes things more

intuitive for you usually like this 5 -

x is a bit to work with a bit weird to

work with on occasion um but if you like

it then just leave it as as

is well now now I've got two fractions

on the left I want the same denominator

so I'm going to multiply my first

fraction on top and bottom by x -

5 then I'll combine those two fractions

over that common denominator giving me

X2 I've got - 5x and plus 4X X so that's

-

x let's get rid of

this so now I actually see that my

numerator factors so let's come over

here so Factor an X Out On Top giving me

x - one and on bottom I've got x - 5 you

can see why factoring is handy because

now I can easily identify my critical

values X can't be five that would make

my denominator zero but X can be 0 or

one in my

numerator to make the numerator

zero so let's cut our number line at

0 1 and

five and see what happens to the left

hand side well if x is five the

denominator is zero so the left hand

side's undefined and if x is 0o one the

numerator is zero so the left hand side

equals

zero and then let's start over here on

the right with our test points let's

plug in a th on the left hand side we

get a positive times a positive over a

positive which simplifies to a positive

let's plug in three here we get positive

* positive over negative that's

negative let's plug in 0. five here we

get positive * negative over negative

which is a

positive and then finally let's plug in

negative 1000 we get a negative * a

negative over a negative which is

negative and we're looking for the

intervals that make the left hand side

positive so that's here and here so

we'll shade those two intervals but not

the end points because they're not

positive they're

zero and we won't shade five because

it's not positive either it's undefined

and so our solution set is 0 to 1 Union

5 to

Infinity okay so there are several

examples solving rational inequalities

common denominators are important

critical values are important so

factoring is important and then the sign

chart very important um but lots of good

examples I hope You' got it down go get

some practice done and good luck