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### Solving Inequalities

- If both sides of an inequality are increased or decreased by the same number, or multiplied or divided by the same positive number, the resulting inequality is equivalent to the original one and has the same solutions.
- If both sides of an inequality are multiplied or divided by the same negative number, the direction of the resulting inequality must be reversed. The new inequality is equivalent to the original one and has the same solutions.
- Use set builder notation to describe the solution set of an inequality.
- The solutions to these inequalities can also be shown by shading the solution set on a number line. Endpoints should be drawn as solid dots if they are part of the solution set, and open circles otherwise.

### Solving Inequalities

- Divide both sides by − 10
- x >− 10
- Recall that whenever you divide inequalities by a negative number, the inequality sign must be flipped,
- therefore, the correct answer is

- Divide both sides by − 25
- x <− 2
- Recall that whenever you divide inequalities by a negative number, the inequality sign must be flipped,
- therefore, the correct answer is

- Subtract 10 from both sides
- − 5x < 40
- Divide both sides by − 5
- x <− 8
- Recall that whenever you divide inequalities by a negative number, the inequality sign must be flipped,
- therefore, the correct answer is

- Subtract 11 from both sides
- − 7x > 21
- Divide both sides by − 7
- x >− 3
- therefore, the correct answer is

- Add 8 to both sides
- − 6x > 48
- Divide both sides by − 6
- x >− 8
- therefore, the correct answer is

- Subtract 14 from both sides
- − 9x >− 54
- Divide both sides by − 9
- x > 6
- therefore, the correct answer is

- Distribute where possible
- [( − 2x + 4)/( − 4)] + 4 − 6x ≤ 6 − x
- Isolate the fraction on the left
- [( − 2x + 4)/( − 4)] ≤ 2 + 5x
- Multiply the entire problem by − 4 in order to eliminate it from the denominator
- − 4( [( − 2x + 4)/( − 4)] ≤ 2 + 5x )
- − 4( [( − 2x + 4)/( − 4)] ) ≤ − 4( 2 + 5x )
- − 2x + 4. − 8 − 20x
- Solve for x
- 12 − 18x
- − [12/18] ≤ x

- Distribute where possible
- [(3x − 12)/( − 5)] + 6 − 12x ≤ 4 − 2x
- Isolate the fraction on the left
- [(3x − 12)/( − 5)] ≤ − 2 + 10x
- Multiply the entire problem by − 5 in order to eliminate it from the denominator
- − 5( [(3x − 12)/( − 5)] ≤ − 2 + 10x )
- − 5( [(3x − 12)/( − 5)] ) ≤ − 5( − 2 + 10x )
- 3x − 1210 − 50x
- Solve for x
- − 22 − 53x
- [( − 22)/( − 53)] ≤ x

- Distribute where possible
- [(3x + 12)/( − 2)] + 4 − 10x6 + 5x
- Isolate the fraction on the left
- [(3x + 12)/( − 2)]2 + 15x
- Multiply the entire problem by − 2 in order to eliminate it from the denominator
- − 2( [(3x + 12)/( − 2)]2 + 15x )
- − 2( [(3x + 12)/( − 2)] ) − 2( 2 + 15x )
- 3x + 12 ≤ − 4 − 30x
- Solve for x
- 16 ≤ − 33x
- [16/( − 33)]x

- Distribute where possible
- [(5(x − 3))/( − 3)] − 12 + 20x10 + 6x
- Isolate the fraction on the left
- [(5(x − 3))/( − 3)]22 − 14x
- Multiply the entire problem by − 3 in order to eliminate it from the denominator
- − 3( [(5(x − 3))/( − 3)]22 − 14x )
- − 3( [(5(x − 3))/( − 3)] ) − 3( 22 − 14x )
- 5x + − 15 ≤ − 66 + 42x
- Solve for x
- 51 ≤ 37x
- [51/37] ≤ x

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Solving Inequalities

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Properties of Inequalities 0:08
- Addition Property
- Example: Using Numbers
- Subtraction Property
- Example: Using Numbers
- Multiplication Properties 1:44
- C>0 (Positive Number)
- Example: Using Numbers
- C<0 (Negative Number)
- Example: Using Numbers
- Division Properties 4:11
- C>0 (Positive Number)
- Example: Using Numbers
- C<0 (Negative Number)
- Example: Using Numbers
- Describing the Solution Set 6:10
- Example: Set Builder Notation
- Example: Graph (Closed Circle)
- Example: Graph (Open Circle)
- Example 1: Solve the Inequality 7:58
- Example 2: Solve the Inequality 9:06
- Example 3: Solve the Inequality 10:10
- Example 4: Solve the Inequality 13:12

### Algebra 2

### Transcription: Solving Inequalities

*Welcome to Educator.com.*0000

*In today's lesson, we are going to be discussing solving inequalities.*0002

*Recall the properties of inequalities: these are needed to solve inequalities.*0008

*The first one is that, if a is greater than b, and you add the same number to both sides, this inequality has the same solution set as the original inequality.*0015

*And you can think of this using numbers to make it clearer.*0030

*If I said that 6 is greater than 5, and then I decided to add 2 to both sides, 8 is greater than 7; so this still holds up.*0035

*You are allowed to add the same number to both sides of an inequality; and that is true for both greater than and less than.*0048

*And it is also true for greater than or equal to and less than or equal to.*0056

*The subtraction property states that, if a is greater than b, and I subtract the same number from both sides,*0064

*then the resulting inequality has the same solution set as the original inequality.*0072

*The same idea as up here: if 6 is greater than 5, and I subtract 3 from both sides, I am going to end up with 3 > 2; that still holds up.*0079

*And that also is valid for less than, and for greater than or equal to and less than or equal to.*0093

*With multiplication properties, if a is greater than b, and a positive number (c > 0), the same positive number,*0104

*is multiplied by both sides of the inequality, then the solution set of the resulting inequality is the same as that of the original inequality.*0113

*If I have 3 > 2, and I multiply both sides by 4, I am going to end up with 12 > 8, which is still valid.*0125

*The same for less than: a < b--you are allowed to multiply both sides of the inequality by the same positive number, without changing the solution set.*0142

*Again, this applies to greater than or equal to and less than or equal to, as well as to strict inequalities.*0152

*The case is different for negative numbers: if c is less than 0 (if you were multiplying both sides of an inequality*0161

*by a negative number), you must reverse the inequality sign.*0167

*If both sides of an inequality are multiplied by the same negative number, the direction of the inequality must be reversed.*0172

*And if you do that, then the resulting inequality has the same solution set as the original inequality.*0179

*If you don't reverse them, the solution set will not necessarily be the same.*0184

*For example, if I have 8 > 4, and I multiply both sides by -2, let's say I didn't reverse the sign:*0190

*then, I am going to end up with -16 > -8; and this is clearly not true.*0205

*So, as soon as I multiply by a negative number, this is incorrect; I immediately have to reverse the direction of the inequality.*0212

*And that will give me -16 < -8, which is valid; and again, the same for greater than or less than or equal to.*0226

*So, multiplying by a positive number, the solution set of the resulting inequality is the same as the original.*0235

*Multiplying by a negative number, you need to reverse the inequality sign in order for the solution set for this inequality to be the same as for the original.*0241

*A similar idea with division: if you are dividing both sides of an inequality by a positive number,*0251

*the resulting inequality has the same solution set as the original.*0259

*That is only if it is for a positive number; so 15 > 20--if I wanted to divide both sides by 5, I could do that.*0264

*And I don't need to do anything to the inequality; I just keep it the same.*0275

*And that is going to give me 3 >...well, I need to start out with an inequality that is valid!*0279

*So, if 20 > 15, and then I divide the same number by both sides...so I am starting out with 20 > 15;*0288

*20 divided by 5 is greater than 15 divided by 5, which is going to give me 4 > 3, which is true.*0305

*The same holds up for less than, and greater than or equal to/less than or equal to.*0314

*Now, if we are dealing with dividing by a negative number, you have to reverse the inequality sign.*0320

*And then, that resulting inequality is the same solution set as the original inequality.*0325

*So, if I have 4 < 6, and I divide both sides by -2, I need to immediately reverse this inequality symbol.*0330

*And then, I am going to get 6 divided by -2, and I am going to end up with -2 > -3; and that is valid.*0344

*If I hadn't reversed this, I would have gotten something that is not valid, or does not have the same solution set as the original.*0354

*So, you need to be very careful, as soon as you multiply or divide by a negative number,*0360

*when you are working with inequalities, that you reverse the inequality symbol.*0366

*There are multiple ways of describing the solution set of an inequality.*0371

*You can either express it as a graph, a set, or simply as an inequality.*0379

*For example, if I came up with the solution set x ≥ 2, I could just leave it as an inequality like that; that is a little less formal.*0386

*I could describe it as a set, using set builder notation.*0396

*And I would put it as follows: what this is saying is "the set of all x, such that x is greater than or equal to 2."*0400

*So, this is set builder notation: the set of all x, such that x is greater than or equal to 2.*0414

*You will see the same things, sometimes, with two dots here; this is more common, but you will see this sometimes; and it means the same thing.*0420

*You can also use a graph on the number line; you can express it as a graph.*0429

*And remember that, if you are saying "greater than or equal to," you are including this number 2 in the solution set.*0436

*So, in that case, you would want to use a closed circle, and then continue on to the right.*0443

*Now, let's say I was going to say x is less than 3.*0450

*In this case, it is a strict inequality; and 3 is not part of the solution set, so I am going to use an open circle.*0458

*So, you can express an inequality's solution set either using set builder notation, using a graph, or less formally, just as a simple inequality.*0465

*Looking at Example 1, we have -3x < -27, so I need to solve that.*0478

*I need to isolate this x; and in order to do that, I am going to need to divide by -3.*0489

*But as soon as I start thinking about dividing by a negative number, I have to reverse the inequality symbol.*0494

*So, instead of less than, this becomes greater than.*0503

*This is going to give me x > 9.*0507

*I can leave it just as an inequality; I can use set notation to show this; I can also graph it on the number line.*0511

*OK, and graphing it on the number line, I am going to use an open circle (since it is a strict inequality)*0532

*and show that x is greater than 9, but that 9 is not part of the solution set.*0538

*Solve -4x - 7 ≥ 9: the first step is to add 7 to both sides, to get -4x ≥ 16.*0547

*Now, to isolate x, I need to divide both sides by -4; and as soon as I do that, I am going to change this direction to less than or equal to.*0561

*16 divided by -4 is -4, so the solution is x ≤ -4.*0570

*Using set notation, and graphing on the number line--multiple ways...-1, -2, -3, -4.*0578

*And this is saying "less than or equal to"; therefore, -4 is going to be included in the solution set, as indicated by a closed circle.*0596

*OK, in Example 3, it is a little bit more complicated; but you just handle it using the same principles.*0611

*When you are dealing with fractions, the best thing to do is to get rid of them first, because they are difficult to deal with.*0621

*So, I am going to multiply both sides of the equation by -9, and (since that is a negative number)*0628

*immediately reverse the inequality symbol, so that you don't forget to do that.*0633

*This is going to be -9 times -2 times 3x - 6; I'll move this over a bit...and that is +(-9), times 4x, divided by -9.*0637

*It is greater than -3 times -9.*0657

*OK, so -9 times -2 is 18, times 3x - 6...these -9's cancel out, and that is going to give me...+4x, is greater than (-3 times -9 is) 27.*0660

*Using the distributive property, I am going to multiply this out; and 18 times 3x is 54x; 18 times -6 is -108; plus 4x, is greater than 27.*0677

*Now, I need to isolate this x; so first, I am going to combine like terms.*0696

*I have 54x + 4x, is 58x, -108 is greater than 27.*0702

*Adding 108 to both sides will give me 58x > 27 + 108, so that is 58x > 135.*0712

*So, x is greater than 135/58; I can leave it like that, or I can use set notation.*0729

*To graph this, you need to figure out...if you divide 135/58, it is a little bit larger than 2; it is approximately 2.3.*0746

*So, I could go ahead and graph that out, as well.*0754

*2 is here; 2.5 is about there; 2.3 is about right here; open circle at 2.3, or actually right over here;*0762

*1 is here; 2 is here; 2.5 is about here; so, 2.3 is going to be right about here; and that is open circle, like that.*0771

*There are three ways to express the solution.*0788

*OK, in Example 4, again, it is a little bit more complex, and there is a fraction, so we are going to get rid of that first.*0794

*Start out by multiplying both sides by -7, which tells me that I immediately need to reverse the sign.*0803

*-7 times -3 times (x - 4), over -7, plus -7 times 3 times (9 - 2x); reverse the inequality symbol--greater than or equal to -7 times (4 - x).*0811

*OK, this cancels; that gives me -3 times (x - 4), and this is -7 times 3, so that is -21, times (9 - 2x), is greater than or equal to...*0843

*-7 times 4; that is -28; -7 times -x is plus 7x.*0860

*So now, I just need to do some more simplification.*0868

*-3 times x is -3x; -3 times -4 is +12; -21 times 9 is -189; -21 times -2x is +42x.*0872

*And this is -28 + 7x; I can't really do anything with that.*0888

*Now, I am going to combine like terms (and I do have some like terms).*0905

*I have a -3x and 42x; combining those, I am going to get 39x; -189 +12 is -177.*0908

*The next step is to isolate the x, so I am going to add 177 to both sides; that is going to give me 177 - 28 + 7x.*0921

*Subtract 7x from both sides: 39x - 7x ≥ 177 - 28.*0934

*Combine like terms to get (39x - 7x is) 32x ≥ 149.*0945

*Now, I am going to divide both sides by 32; and it is a positive number, so it just becomes x ≥ 149/32.*0956

*Using set notation, x is greater than or equal to 149/32.*0970

*You could also graph this; this is approximately equal to 4.7, so 0, 1, 2, 3, 4, and 5;*0977

*it is going to be a little over halfway between 4 and 5; closed circle; and graph it.*0989

*OK, so the first step was to eliminate the fraction by multiplying both sides of the equation by -7 and reversing the inequality symbol.*0996

*Once that was done, we are using the distributive property to multiply everything out and get rid of the parentheses.*1005

*Then, we are using the addition and subtraction principles and combining like terms to simplify this inequality.*1012

*And the result was x ≥ 149/32.*1021

*That concludes this lesson on solving inequalities at Educator.com; see you again!*1030

1 answer

Last reply by: Manfred Berger

Tue May 28, 2013 6:39 AM

Post by Angela Belue on September 16, 2011

what about Two-Variable Inequalities?

1 answer

Last reply by: Rafael Wang

Sun Mar 12, 2017 9:36 PM

Post by Ashraf Saeed on November 14, 2010

haha 4:15