6. Solving Inequalities
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Chapter 6
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.
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| | 15 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Inequalities
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.
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.
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b What is the minimum total number of tickets the club needs to sell to reach their goal?
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Discussion
Properties of Inequalities
Not all inequalities are expressed in the form x<a, x>a, x≤a, or x≥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.
Proof
Addition Property of InequalityThe 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.
Using the biconditional statement, the last inequality can be rewritten.
x<y⇔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
(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.
Proof
Subtraction Property of InequalityThe 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.
From the biconditional statement, the last inequality can be rewritten.
x<y⇔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
(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=0. Then, the equivalent inequalities can be written depending on the sign of z.
- If x<y and z>0, then xz<yz.
- If x<y and z<0, then xz>yz.
Proof
Multiplication Property of InequalityThe 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 xy>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 xz<yz.
- If x<y and z<0, then xz>yz.
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⇔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>0z(y−andz>0⇓x)>0
Now, the second part of this conditional statement can be rewritten using the Distributive Property. z(y−x)>0⇔zy−zx>0
From the first property, it can be said that zy−zx>0 if and only if zx<zy. Additionally, because x<y, the conditional statement has been proven. If x<y and z>0, then zx<zy.
When z Is Less Than 0
Again, because x<y, the following statement is valid.x<y⇔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=-ab
-zy−(-zx)>0
SubNeg
a−(-b)=a+b
-zy+zx>0
AddIneq
LHS+zy>RHS+zy
zx>zy
If x<y and z<0, then zx>zy.
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=0. Then, the equivalent inequalities can be written depending on the sign of z.
- If x<y and z>0, then zx<zy.
- If x<y and z<0, then zx>zy.
Proof
Division Property of InequalityThe 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 zx 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 zx<zy.
- If x<y and z<0, then zx>zy.
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⇔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>0zy−xandz>0⇓>0
Now, the second part of this conditional statement can be rewritten. zy−x>0⇔zy−zx>0
By using the first property, it can be said that zx is less than zy. Additionally, because x<y, the property has been proven. If x<y and z>0, then zx<zy.
z<0
Again, because x<y, the following statement is valid.x<y⇔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-zy−xand-z>0⇓>0
Now, the second part of this statement can be rewritten.
-zy−x>0
▼
Simplify
MoveNegDenomToNum
Put minus sign in numerator
z-(y−x)>0
DistrNegSignSwap
-(b−a)=a−b
zx−y>0
WriteDiffFrac
Write as a difference of fractions
zx−zy>0
AddIneq
LHS+zy>RHS+zy
zx>zy
If x<y and z<0, then zx>zy.
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.
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.
3x+21>6−2x
Three Properties of Inequalities will be used to solve this inequality.
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.
4s−132≤s+174
The number of seats b in the balcony is defined by another inequality.
143−3b>b−17
a Find the simplest form of the inequality for the number of ground floor seats.
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b Find the simplest form of the inequality for the number of seats on the balcony.
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c What is the maximum number of seats in the theater hall according to the given inequalities?
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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.
4s−132≤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.
Next, add 132 to both sides of the inequality by using the Addition Property of Inequality.
Finally, divide both sides of the inequality by 3 to get the variable alone by using the Division Property of Inequality.
The simplest form of the given inequality is obtained.
4s−132≤s+174⇕s≤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.
143−3b>b−17
Once again, use the properties of inequalities to simplify it. Start by adding 3b to both sides of the inequality by using the Addition Property of Inequality.
Now, add 17 to both sides of the inequality.
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.
The simplest form of the given inequality is found.
143−3b>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 Floors≤102 and Balconyb<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.
Lastly, the solution set of the inequality can be represented using set-builder notation. 
2x−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⇒-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. 2x−3<5
x<4
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.
-2h+30≤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.
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b Find the solution set of the inequality that represents the score of the literature and language class.
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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.
-2h+30≤h−315
First, subtract 30 from both sides of the inequality using the Subtraction Property of Inequality.
Next, subtract h from both sides of the inequality.
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.
-3h≤-345
DivNegIneq
Divide by -3 and flip inequality sign
-3-3h≥-3-345
CalcQuot
Calculate quotient
h≥115
h≥115⇓{h∣h≥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
55(s−14)>5245
CancelCommonFac
Cancel out common factors
55(s−14)>5245
CalcQuot
Calculate quotient
s−14>49
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. 1
Determine the Type of Inequality
expand_more 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
2
Determine the Solution Set and the Boundary Point
expand_more 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.
Therefore, the boundary point is 6 and the solution set corresponds to all real numbers less than 6.
3
Draw the Boundary Point on the Number Line
expand_more 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.
4
Shade the Rest of the Solution Set
expand_more 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.
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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.
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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.
The number of tickets needs to be an integer. The minimum integer value of x greater than or equal to 24.67 is 25. Now, add this number to 42 to find the total number of tickets.
42⋅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 ≤. This will represent the left hand side of the inequality.
400≤
Now, let x be the number of additional tickets to be sold after 42 tickets. Then 6x will represent the amount that will be earned. This means that the total amount will be 252+6x. 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≤252+6x⇕252+6x≥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.
Finally, divide both sides of the inequality by 6 by using the Division Property.
148≤6x
DivIneq
LHS/6≤RHS/6
6148≤66x
CalcQuot
Calculate quotient
24.666666…≤x
RoundDec
Round to 2 decimal place(s)
24.67≤x
42+25=67
The club needs to sell 67 tickets in total to earn at least $400.
Write the given word sentences as an inequality.
Twice a number b is at most 18.
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A number d minus 3 is greater than or equal to -5.
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We will represent the given sentence with an inequality. Let's start by examining each phrase in the sentence.
Twice a numberb is at most 18
We can write Twice a number b
as a product of 2 and b.
Twice a numberb
2b
Next, we will write the phrase at most
as an inequality symbol. Let's review the inequality symbols and their matching word phrases.
| Symbol | Key Phrases |
|---|---|
| < | is less than is fewer than |
| > | is greater than is more than |
| ≤ | is less than or equal to is at most is no more than |
| ≥ | is greater than or equal to is at least is no less than |
We can write is at most
as ≤. Let's use this information and finally rewrite the whole sentence as an inequality.
ccc
Twice a numberb & is at most & 18
2b & ≤ & 18
Once again we will write the following sentence as an inequality.
A numberd minus 3 is greater than equal to -5.
Note that minus
represents subtraction.
d - 3
We obtained the left hand side of the inequality. The right hand side will be -5.
d - 3 ? -5
Finally, we need to determine the inequality symbol between them. Let's have a look at the table showing the inequality symbols in Part A.
| Symbol | Key Phrases |
|---|---|
| < | is less than is fewer than |
| > | is greater than is more than |
| ≤ | is less than or equal to is at most is no more than |
| ≥ | is greater than or equal to is at least is no less than |
We will use the sign ≥ between the expressions. A numberd minus 3 is greater than equal to -5 ⇓ d - 3 ≥ -5
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Decide whether the given value is a solution of the inequality.
m−5 > 7; m=4
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2n+6≤ -1; n=-14
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We are asked to check whether the given value is a solution of the inequality or not. m-5 > 7; m= 4 We can determine this by substituting the given value into the inequality. 4-5 > 7 Now, let's calculate the left hand side and see whether we get a true statement or not.
Notice that the obtained inequality is not true because -1 is not greater than 7. This means that m= 4 is not a solution for the given inequality.
Once again we will check if n=-14 is a solution for the given inequality. Let's substitute n=-14 into the inequality and evaluate it.
Great! We ended with a true statement because -1 is equal to -1. It satisfies the equality part of the symbol greater than or equal to
. This means that n= -14 is a solution of the given inequality.
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Find the solution set of the given inequalities.
x−4≤19
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y+12>-5
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We will solve the given inequality. x-4 ≤ 19 Recall that we can solve it by isolating the variable x on one side of the inequality. Let's add 4 to both sides of the inequality to isolate x at the left hand side by using the Addition Property of Inequality.
Great! Now we will write the obtained solution using set-builder notation. x ≤ 23 ⇓ {x | x ≤ 23 }
Once again we will find the solution set of the given inequality. y+12 > -5 This time we will use the Subtraction Property of Inequality to isolate the variable y. Let's subtract 12 from both sides of the inequality.
Now that we expressed the solution, we will use set-builder notation to write it as a solution set. y > -17 ⇓ {y | y > -17 }
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Find the solution set of the given inequalities.
3m<-5
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-2n≥14
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We will solve the given inequality. m/3 < -5 Recall that we can solve it by isolating the variable m on one side of the inequality. Let's multiply both sides of the inequality by 3 to isolate m at the left hand side by using the Multiplication Property of Inequality.
Great! Now we will write the obtained solution using set-builder notation. m ≤ -15 ⇓ {m | m ≤ -15 }
Once again we will find the solution set of the given inequality. -2n ≥ 14 Notice that we need to divide both side of the inequality by -2 to isolate the variable n. According to the Division Property of Inequality, we should reverse the inequality sign. Let's do it!
Now that we expressed the solution, we will use set-builder notation to write it as a solution set. n ≤ -7 ⇓ {n | n ≤ -7 }
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3x+17<56
Choose the graph that represents the solution set of the given inequality.
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3y≥1
Choose the graph that represents the solution set of the given inequality.
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We will show the solution set of the given inequality on a number line. 3x+17 < 56 Let's first solve the inequality. Start by subtracting 17 from both sides of the inequality by using the Subtraction Property.
Next, let's divide both sides of the obtained inequality by 3.
We found the solution. Since the inequality is strict, the solution set does not include the boundary value 13. This means that we will show 13 with a blank circle on the number line.
Then, we will choose a test point to check which side of the number line we should shade. We can take x= 11.
Notice that 11 is a solution of our inequality. This means that we will shade the left hand side of the number line since 11 stays at the left hand side of 13.
We obtained the graph at the first option. The answer is Graph I.
Once again we will graph the solution set of the given inequality. Let's first solve it by using the Multiplication Property.
Notice that the inequaIty is non-strict. This means that the solution set includes the boundary value.
Great! Next, let's choose a test point to check which side of the number line we should shade. We can take x=0. 0 ≥ 3 * Since x=0 does not satisfy the inequality, we will shade the side that does not include 0 which is the right hand side of the number line.
The obtained number line is the same as with the fourth option Graph IV. Notice that the Graph I is also includes the numbers greater than or equal to 3 but it just shows the integers. Since there is no restriction about the solution set, we need to consider all real numbers which is the reason to eliminate Graph I.
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Solving Inequalities
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