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Pre-Algebra View details

6. Solving Inequalities

Lesson

Exercises

Tests

eCourses /

Pre-Algebra /

Chapter 6

Solving Inequalities

This lesson delves into the methods for solving inequalities, focusing on the properties of inequalities, real numbers, and boundary points. It explains how these properties can be used to isolate variables and find the solution set. For example, the Addition and Subtraction Properties of Inequalities are discussed as tools for simplifying inequalities. The lesson also provides practical scenarios, such as determining the number of seats in a theater or the height requirement for joining a club, to illustrate the application of these mathematical concepts. It emphasizes that inequalities can have multiple or even infinitely many solutions, which can be represented on a number line.

Lesson Settings & Tools

	15 Theory slides
	9 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Image Credits

Slide of 15

Inequalities can be solved like equations. The key step is to isolate the variable on one side of the inequality. Inequalities may have multiple, or even infinitely many, solutions. In this lesson several examples are introduced showing how to use the properties of inequalities to solve them.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Number Line
Inverse Operations
Inequality
Strict and Non-strict Inequalities
Solution Set of an Inequality
Writing and Graphing Inequalities

Challenge

Tickets for a Drama: Making Money

Jordan is a member of the drama club at her school. The club plans to stage the play The Little Frog in Town this weekend. They charge $$ 6$ per ticket.

a The club hopes to earn at least

$ 400

from the play. If the club has already sold

42

tickets, write an inequality to represent the number of tickets they still need to sell.

b What is the minimum total number of tickets the club needs to sell to reach their goal?

Discussion

Properties of Inequalities

Not all inequalities are expressed in the form $x < a,$ $x > a,$ $x \leq a,$ or $x \geq a .$ Yet, through inverse operations and the Properties of Inequalities, any inequality can be simplified to one of the mentioned forms. Consider the Addition and Subtraction Properties of Inequalities.

Rule

Addition Property of Inequality

Adding the same number to both sides of an inequality generates an equivalent inequality. This equivalent inequality will have the same solution set and the inequality sign remains the same. Let $x,$ $y,$ and $z$ be real numbers such that $x < y .$ Then, the following conditional statement holds true.

If $x < y,$ then $x + z < y + z .$

This property holds for the other types of inequalities.

Proof

Addition Property of Inequality

The case when

x < y

will be proven. The remaining cases can be proven similarly. Before starting the proof, the following biconditional statement needs to be considered.

x < y \Leftrightarrow y - x > 0

Now, the Identity Property of Addition can be applied to the second part of the statement.

y - x > 0

IdPropAdd

Identity Property of Addition

y - x + 0 > 0

Rewrite $0$ as $z - z$

y - x + z - z > 0

CommutativePropAdd

Commutative Property of Addition

y + z - x - z > 0

$- a - b = - (a + b)$

y + z - (x + z) > 0

AddPar

Add parentheses

(y + z) - (x + z) > 0

Using the biconditional statement, the last inequality can be rewritten.

(y + z) - (x + z) > 0 ⇕ x + z < y + z

Finally, because

x < y,

the property is obtained.

If $x < y,$ then $x + z < y + z .$

Discussion

Subtraction Property of Inequality

Subtracting the same number from both sides of an inequality produces an equivalent inequality. The solution set and inequality sign of this equivalent inequality does not change. Let $x,$ $y,$ and $z$ be real numbers such that $x < y .$ Then, the following conditional statement holds true.

If $x < y,$ then $x - z < y - z .$

This property holds for the other types of inequalities.

Proof

Subtraction Property of Inequality

The case when

x < y

will be proven. The other cases can be proven using a similar reasoning. Consider the biconditional statement before beginning the proof.

x < y \Leftrightarrow y - x > 0

This property can be proven using the Additive Inverse of

z,

which is

- z .

Now, the Identity Property of Addition can be applied to the second part of the statement.

y - x > 0

IdPropAdd

Identity Property of Addition

y - x + 0 > 0

Rewrite $0$ as $(- z) - (- z)$

y - x + (- z - (- z)) > 0

CommutativePropAdd

Commutative Property of Addition

y + (- z - (- z)) - x > 0

RemovePar

Remove parentheses

y - z - (- z) - x > 0

$- a - b = - (a + b)$

y - z - (- z + x) > 0

CommutativePropAdd

Commutative Property of Addition

y - z - (x - z) > 0

AddPar

Add parentheses

(y - z) - (x - z) > 0

From the biconditional statement, the last inequality can be rewritten.

(y - z) - (x - z) > 0 ⇕ x - z < y - z

Finally, because

x < y,

the property has been proven.

If $x < y,$ then $x - z < y - z .$

Discussion

Multiplication Property of Inequality

Multiplying both sides of an inequality by a nonzero real number $z$ produces an equivalent inequality. The following conditions about $z$ need to be considered when applying this property.

Positive $z$	If $z$ is positive, the inequality sign remains the same.
Negative $z$	If $z$ is negative, the inequality sign needs to be reversed to produce an equivalent inequality.

For example, let $x,$ $y,$ and $z$ be real numbers such that $x < y$ and $z \neq = 0 .$ Then, the equivalent inequalities can be written depending on the sign of $z .$

If $x < y$ and $z > 0,$ then $x z < y z .$
If $x < y$ and $z < 0,$ then $x z > y z .$

This property holds for the other types of inequalities.

Proof

Multiplication Property of Inequality

The case when $x < y$ will be proven. The remaining cases can be proven following a similar reasoning. Before starting the proof, the following properties of real numbers need to be considered.

$x < y$ if and only if $y - x > 0 .$
If $x$ and $y$ are positive, then $x y > 0 .$
If $z$ is negative, then $- z$ is positive.

Using these properties, the following conditional statements can be proven.

If $x < y$ and $z > 0,$ then $x z < y z .$
If $x < y$ and $z < 0,$ then $x z > y z .$

Each conditional statement will be analyzed separately.

When $z$ Is Greater Than $0$

It is given that

x < y,

then using the first property, it is known that

y - x

is greater than

0 .

x < y \Leftrightarrow y - x > 0

Furthermore, because

z > 0,

from the second property, it can be stated that the product of

z

and

y - x

is also greater than

0 .

y - x > 0 z (y - and z > 0 ⇓ x) > 0

Now, the second part of this conditional statement can be rewritten using the Distributive Property.

z (y - x) > 0 \Leftrightarrow z y - z x > 0

From the first property, it can be said that

z y - z x > 0

if and only if

z x < z y .

Additionally, because

x < y,

the conditional statement has been proven.

x < y

and

z > 0,

then

z x < z y .

When $z$ Is Less Than $0$

Again, because

x < y,

the following statement is valid.

x < y \Leftrightarrow y - x > 0

Additionally, since

z < 0,

from the third property it follows that

- z

is positive. Moreover, the product of

- z

and

y - z

will be positive.

y - x > 0 - z (y - and - z > 0 ⇓ x) > 0

Now,

- z

can be distributed in the second part of the statement.

- z (y - x) > 0

▼

Simplify

Distr

Distribute $(- z)$

(- z) y - (- z) x > 0

MultNegPos

$(- a) b = - a b$

- z y - (- z x) > 0

SubNeg

$a - (- b) = a + b$

- z y + z x > 0

AddIneq

$LHS + z y > RHS + z y$

z x > z y

Finally, because

x < y,

the property has been proven.

If $x < y$ and $z < 0,$ then $z x > z y .$

Discussion

Division Property of Inequality

Dividing both sides of an inequality by a nonzero real number $z$ produces an equivalent inequality. However, the following conditions need to be considered.

Positive $z$	If $z$ is positive, the inequality sign remains the same.
Negative $z$	If $z$ is negative, the inequality sign needs to be reversed to produce an equivalent inequality.

For example, let $x,$ $y,$ and $z$ be real numbers such that $x < y$ and $z \neq = 0 .$ Then, the equivalent inequalities can be written depending on the sign of $z .$

If $x < y$ and $z > 0,$ then $\frac{x}{z} < \frac{y}{z} .$
If $x < y$ and $z < 0,$ then $\frac{x}{z} > \frac{y}{z} .$

This property holds for the other types of inequalities.

Proof

Division Property of Inequality

The case when $x < y$ will be proven. The remaining cases can be proven following a similar reasoning. Before starting the proof, the following properties of real numbers need to be considered.

$x < y$ if and only if $y - x$ is positive.
If $x$ and $z$ are positive, then $\frac{x}{z}$ is also positive.
If $z$ is negative, then $- z$ is positive.

Using these properties, the following conditional statements can be proven.

If $x < y$ and $z > 0,$ then $\frac{x}{z} < \frac{y}{z} .$
If $x < y$ and $z < 0,$ then $\frac{x}{z} > \frac{y}{z} .$

Each case will be analyzed separately.

$z > 0$

It is given that

x < y,

then using the first property, it is known that

y - x

is greater than

0 .

x < y \Leftrightarrow y - x > 0

Furthermore, because

z > 0,

from the second property, it can be stated that

y - x

divided by

z

is also greater than

0 .

y - x > 0 \frac{y - x}{z} and z > 0 ⇓ > 0

Now, the second part of this conditional statement can be rewritten.

\frac{y - x}{z} > 0 \Leftrightarrow \frac{y}{z} - \frac{x}{z} > 0

By using the first property, it can be said that

\frac{x}{z}

is less than

\frac{y}{z} .

Additionally, because

x < y,

the property has been proven.

If $x < y$ and $z > 0,$ then $\frac{x}{z} < \frac{y}{z} .$

$z < 0$

Again, because

x < y,

the following statement is valid.

x < y \Leftrightarrow y - x > 0

Additionally, since

z < 0,

from the third property, it follows that

- z

is positive. Moreover, the quotient of

y - x

and

- z

will be positive.

y - x > 0 \frac{y - x}{- z} and - z > 0 ⇓ > 0

Now, the second part of this statement can be rewritten.

\frac{y - x}{- z} > 0

▼

Simplify

MoveNegDenomToNum

Put minus sign in numerator

\frac{- ( y - x )}{z} > 0

DistrNegSignSwap

$- (b - a) = a - b$

\frac{x - y}{z} > 0

WriteDiffFrac

Write as a difference of fractions

\frac{x}{z} - \frac{y}{z} > 0

AddIneq

$LHS + \frac{y}{z} > RHS + \frac{y}{z}$

\frac{x}{z} > \frac{y}{z}

Finally, because

x < y,

the property has been obtained.

If $x < y$ and $z < 0,$ then $\frac{x}{z} > \frac{y}{z} .$

Pop Quiz

Identifying the Appropriate Property to Solve the Inequality

In the applet, determine the property used to isolate the variable on one side of the given inequality as shown.

An applet showing different inequalities and its equivalent inequality that results of applying one of the properties of inequalities

Discussion

Equivalent Inequalities

Two or more inequalities are equivalent inequalities if they have the same solution set. Similar to equivalent equations, applying Properties of Inequality to an inequality produces an equivalent inequality. Consider the following example.

3 x + 21 > 6 - 2 x

Three Properties of Inequalities will be used to solve this inequality.

3 x + 21 > 6 - 2 x

AddIneq

$LHS + 2 x > RHS + 2 x$

5 x + 21 > 6

SubIneq

$LHS - 21 > RHS - 21$

5 x > - 15

DivIneq

$LHS / 5 > RHS / 5$

x > - 3

The solution set of every inequality created in this process is

x > - 3 .

The Properties of Inequalities produced three equivalent inequalities until the solution set was found.

Example

Number of Seats in the Theater Hall

The play The Little Frog in Town received rave reviews from the audience after its first showing. The club decides to put on this play again with one major difference — they will hold it in the greatest theater hall their city has to offer! Now, the drama club needs to know the number of seats in the hall to print new tickets.

The hall consists of two floors, the ground floor and the balcony. The number of seats

s

on the ground floor is defined by the following inequality.

4 s - 132 \leq s + 174

The number of seats

b

in the balcony is defined by another inequality.

143 - 3 b > b - 17

a Find the simplest form of the inequality for the number of ground floor seats.

b Find the simplest form of the inequality for the number of seats on the balcony.

c What is the maximum number of seats in the theater hall according to the given inequalities?

Hint

a Use the properties of inequalities to simplify the given inequality.

b Use the properties of inequalities to simplify the given inequality.

c Use the simplified inequalities from Part A and Part B.

Solution

a Start by examining the given inequality that represents the number of seats on the ground floor.

4 s - 132 \leq s + 174

Now, use the properties of inequalities to simplify this inequality. Notice that the variable

s

appears on both sides of the inequality. First, subtract

s

from both sides of the inequality by using the Subtraction Property of Inequality.

4 s - 132 \leq s + 174

SubIneq

$LHS - s \leq RHS - s$

4 s - 132 - s \leq s + 174 - s

SubTerms

Subtract terms

3 s - 132 \leq 174

Next, add

132

to both sides of the inequality by using the Addition Property of Inequality.

3 s - 132 \leq 174

AddIneq

$LHS + 132 \leq RHS + 132$

3 s - 132 + 132 \leq 174 + 132

AddTerms

Add terms

3 s \leq 306

Finally, divide both sides of the inequality by

3

to get the variable alone by using the Division Property of Inequality.

3 s \leq 306

DivIneq

$LHS / 3 \leq RHS / 3$

\frac{3 s}{3} \leq \frac{3 0 6}{3}

CalcQuot

Calculate quotient

s \leq 102

The simplest form of the given inequality is obtained.

4 s - 132 \leq s + 174 ⇕ s \leq 102

This means that the number of seats on the ground floor is less than or equal to

102 .

In other words, there are at most

102

seats on the ground floor.

b This time, the following inequality will be simplified.

143 - 3 b > b - 17

Once again, use the properties of inequalities to simplify it. Start by adding

3 b

to both sides of the inequality by using the Addition Property of Inequality.

143 - 3 b > b - 17

AddIneq

$LHS + 3 b > RHS + 3 b$

143 - 3 b + 3 b > b - 17 + 3 b

AddTerms

Add terms

143 > 4 b - 17

Now, add

17

to both sides of the inequality.

143 > 4 b - 17

AddIneq

$LHS + 17 > RHS + 17$

143 + 17 > 4 b - 17 + 17

AddTerms

Add terms

160 > 4 b

As the last step, divide both sides of the inequality by

4 .

Note that since

4

is a positive number, the inequality sign stays the same. In the case that the divisor is a negative number, the inequality sign should be reversed.

160 > 4 b

DivIneq

$LHS / 4 > RHS / 4$

\frac{1 6 0}{4} > \frac{4 b}{4}

CalcQuot

Calculate quotient

40 > b

The simplest form of the given inequality is found.

143 - 3 b > b - 17 ⇕ 40 > b or b < 40

This means that

40

is greater than the number of seats on the balcony. In other words there are at most

39

seats on the balcony.

c Recall the simplified forms of the inequalities in Part A and Part B.

Ground Floor s \leq 102 and Balcony b < 40

According to these inequalities, there are at most

102

seats on the ground floor and

39

seats on the balcony. Now, add these number of seats to find the maximum number of seats in the theater hall.

102 + 39 = 141

The maximum number of seats in the theater hall is

141 .

Discussion

Solution Set of an Inequality

A solution of an inequality is any value of the variable that makes the inequality true. As an example, consider the following inequality.

2 x - 3 < 5

Notice that if

0

is substituted for

x

in the inequality, the inequality holds true. Therefore, it can be said that

0

is a solution to the given inequality.

2 (0) - 3 < ? 5 \Rightarrow - 3 < 5 ✓

However, this is not the only value that makes the inequality true. There are other

x -

values like

1

and

2

that make it true. The set of all possible values that satisfy an inequality is the solution set of an inequality. The solution set can be determined by applying the Properties of Inequalities to isolate the variable on one side of the inequality.

2 x - 3 < 5

▼

Solve for

x

AddIneq

$LHS + 3 < RHS + 3$

2 x < 5 + 3

AddTerms

Add terms

2 x < 8

DivIneq

$LHS / 2 < RHS / 2$

x < \frac{8}{2}

CalcQuot

Calculate quotient

x < 4

Lastly, the solution set of the inequality can be represented using set-builder notation.

\underline{Solution Set} {x ∣ x < 4}

It is worth noting that the solution set of a linear inequality in one variable can also be represented using a number line.

Example

Theater Auditions

After watching excellent theater performances put on by the drama club, so many students want to join it. Several characteristics are highly sought after for new recruits.

For instance, at the moment, the drama club does not have many tall members. For this recruiting year, they have defined a height to consider for new recruits

h

with the following inequality.

- 2 h + 30 \leq h - 315

More importantly, the new recruit needs to get a great score

s

in their Literature & Language class. The score is defined with the following inequality.

5 (s - 14) > 245

a Find the solution set of the inequality for the appropriate height of a member.

b Find the solution set of the inequality that represents the score of the literature and language class.

Hint

a Isolate the variable

h

by using the properties of inequalities.

b Isolate the variable

s

by using the properties of inequalities.

Solution

a The solution set of an inequality can be found by isolating the variable on one side of the inequality. Simplify the given inequality by using the properties of inequalities to isolate the variable

h .

- 2 h + 30 \leq h - 315

First, subtract

30

from both sides of the inequality using the Subtraction Property of Inequality.

- 2 h + 30 \leq h - 315

SubIneq

$LHS - 30 \leq RHS - 30$

- 2 h + 30 - 30 \leq h - 315 - 30

SubTerms

Subtract terms

- 2 h \leq h - 345

Next, subtract

h

from both sides of the inequality.

- 2 h \leq h - 345

SubIneq

$LHS - h \leq RHS - h$

- 2 h - h \leq h - 345 - h

SubTerms

Subtract terms

- 3 h \leq - 345

Now, divide both sides of the obtained inequality by

- 3

to isolate the variable

h .

Note that according to the Division Property of Inequality, the inequality sign should be reversed since the divisor

- 3

is a negative number.

- 3 h \leq - 345

DivNegIneq

Divide by $- 3$ and flip inequality sign

\frac{- 3 h}{- 3} \geq \frac{- 3 4 5}{- 3}

CalcQuot

Calculate quotient

h \geq 115

Lastly, rewrite this inequality by using the set-builder notation to represent the answer as a solution set.

h \geq 115 ⇓ {h ∣ h \geq 115}

This means that this year the drama club is looking for recruits with a height greater than or equal to

115

centimeters.

b Here, the task is to find the solution set of the following inequality.

5 (s - 14) > 245

Once again, proceed in the same way as Part A. Isolate the variable

s

by using the properties of inequalities. That can be done by dividing both sides of the inequality by

5

to get rid of the factor in front of the parentheses.

5 (s - 14) > 245

DivIneq

$LHS / 5 > RHS / 5$

\frac{5 ( s - 1 4 )}{5} > \frac{2 4 5}{5}

CancelCommonFac

Cancel out common factors

\frac{5 ( s - 1 4 )}{5} > \frac{2 4 5}{5}

CalcQuot

Calculate quotient

s - 14 > 49

Next, add

14

to both sides of the inequality to isolate the variable

s .

s - 14 > 49

AddIneq

$LHS + 14 > RHS + 14$

s - 14 + 14 > 49 + 14

AddTerms

Add terms

s > 63

Finally, write the simplified inequality in the set-builder notation.

s > 63 ⇓ {s ∣ s > 63}

This means that if someone wants to join the drama club, they needs to get a score greater than

63 .

Pop Quiz

Solving Inequalities

In the applet, there are given various different inequalities. Select the correct solution set for the given inequality.

Discussion

Graphing an Inequality on a Number Line

A number line can be used to represent the solution set of an inequality that has one variable. To graph such an inequality, first, determine its type. If it is a strict inequality, then an open boundary point is drawn. Otherwise, a closed boundary point is drawn. Then, the rest of the solution set is shaded accordingly. Consider the following inequality.

x + 2 < 8

The following four steps act as a guide in graphing the given inequality.

Determine the Type of Inequality

The first step is determining if the inequality is strict or non-strict. In this case, the given inequality is strict because the inequality symbol is

< .

Strict x + 2 < 8

Determine the Solution Set and the Boundary Point

Next, the solution set and the boundary point of the inequality need to be found. This can be done by solving the inequality using the Properties of Inequalities.

x + 2 < 8

SubIneq

$LHS - 2 < RHS - 2$

x < 6

Therefore, the boundary point is

6

and the solution set corresponds to all real numbers less than

6 .

Draw the Boundary Point on the Number Line

Here, a circle representing the boundary point is drawn on the number line. If the inequality is strict, the circle is open. If the inequality is non-strict, the circle is closed. For this example, the inequality is strict, and the boundary point is $6 .$ Then, an open circle will be drawn on the number line on the number $6 .$

Shade the Rest of the Solution Set

Finally, the rest of the solution set will be shaded by drawing an arrow that goes along the solution set and starts on the boundary point. For this situation, the solution set corresponds to all numbers less than $6,$ which means the arrow will be along the left of the boundary point.

It is worth mentioning that the graph of inequalities whose solution sets are all the real numbers are represented with bidirectional arrows that cover all the number line.

Example

Showing the Duration of the Plays on a Number Line

The drama club makes a survey of several audiences about the plays presented during the school year. Then they discussed these surveys' results at the end-of-year meeting.

This year all plays were at least $100$ minutes long. Since the audiences found the duration of the plays a bit too long according to the survey results, the club takes a decision to keep the next year's plays less than $90$ minutes in length.

a Write an inequality for the duration

d

of the next year's plays.

b Choose the graph that describes the solution set of the inequality written in Part A. This will be a way to present the data.

Number lines representing the possible solution sets of the inequality

Hint

a Use a strict inequality to represent the phrase less than.

b The solution set includes real numbers, not only integers.

Solution

a The club decides to keep the duration of the next year's plays below

90

minutes. Let

d

represent the duration of the plays. Note that less than represent a strict inequality. This means that the inequality sign will be

< .

With this in mind, write an inequality to show

d

is less than

90 .

d < 90

Notice that it can also be written by using the phrasegreater than. In other words, saying

d

is less than

90

is the same as saying

90

is greater than the duration of the plays

d

d < 90 ⇕ 90 > d

b Now, the graph of the inequality obtained in Part A can be drawn.

d < 90

Since the inequality is strict,

90

is not included in the solution set. This means that

90

will be shown with an open circle in the graph. This already eliminates options II and III. Also, notice that the solution set needs to include real numbers, not just integers, because time is a continuous quantity.

Solution Set {d ∣ d is a real number less than 90}

The option IV does not make sense for this solution set because it shows only integers. However, Graph I shows exactly the real numbers less than

90 .

This means that the answer is Graph I.

Closure

Finding the Number of Tickets

At the beginning of the lesson, it was given that the drama club charges

$ 6

per ticket and

42

tickets have been already sold. First, calculate the amount of money earned from these

42

tickets.

42 \cdot 6 = 252

The club earns

$ 252

from selling

42

tickets. Recall that the aim is to earn at least

$ 400

from this play. Note that at least can be represented using the inequality symbol less than or equal to

\leq .

This will represent the left hand side of the inequality.

400 \leq

Now, let

x

be the number of additional tickets to be sold after

42

tickets. Then

6 x

will represent the amount that will be earned. This means that the total amount will be

252 + 6 x .

Recall that this amount needs to be greater than or equal to

400

or in other words,

400

needs to be less than or equal to that amount. With this in mind, now complete the other side of the inequality.

400 \leq 252 + 6 x ⇕ 252 + 6 x \geq 400

Now, solve the obtained inequality to find the minimum number of tickets sold. As before, use the properties of inequalities to isolate the variable

x .

Start by subtracting

252

from both sides of the inequality by using the Subtraction Property.

400 \leq 252 + 6 x

SubIneq

$LHS - 252 \leq RHS - 252$

400 - 252252 + 6 x - 252

SubTerms

Subtract terms

148 \leq 6 x

Finally, divide both sides of the inequality by

6

by using the Division Property.

148 \leq 6 x

DivIneq

$LHS / 6 \leq RHS / 6$

\frac{1 4 8}{6} \leq \frac{6 x}{6}

CalcQuot

Calculate quotient

24.666666 \dots \leq x

RoundDec

Round to $2$ decimal place(s)

24.67 \leq x

The number of tickets needs to be an integer. The minimum integer value of

x

greater than or equal to

24.67

25 .

Now, add this number to

42

to find the total number of tickets.

42 + 25 = 67

The club needs to sell

67

tickets in total to earn at least

$ 400 .

Level 1

Level 2

Level 3

Solving Inequalities

Exercises

Consider the following inequality.

m x + k < m x + ℓ

If it is true for all real values of

x,

what can be said for the following inequality?

m x + k > m x + ℓ

We are told that mx+k < mx+l is true for all real values of x. Notice that the variable term mx is seen on both sides of the inequality. Let's combine like terms using the Subtraction Property of Inequality.

The solution tells us that k is less than l. This means that the variable term has no effect on the solution, which depends only on the relationship between k and l. k < l is always true Now, let's simplify the second inequality as well.

In the beginning, we found that k < l is always true. Therefore, k cannot be greater than l. As a result, we can say that mx+k>mx+l is a false statement for all real values of x.

Solving Inequalities

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Solving Inequalities

Catch-Up and Review

Proof

Proof

Proof

When z Is Greater Than 0

When z Is Less Than 0

Proof

z>0

z<0

Hint

Solution

Hint

Solution

Hint

Solution

Solving Inequalities

Recommended exercises

When $z$ Is Greater Than $0$

When $z$ Is Less Than $0$

$z > 0$

$z < 0$