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Lecture Comments (3)

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?

0 answers

Post by Ashraf Saeed on November 14, 2010

haha 4:15

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

Solve − 10x > 100
  • 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
x < − 10
Solve − 25x < 50
  • 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
x > − 2
Solve − 5x + 10 < 50
  • 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
x > − 8
Solve − 7x + 11 > 32
  • Subtract 11 from both sides
  • − 7x > 21
  • Divide both sides by − 7
  • x > − 3
  • Recall that whenever you divide inequalities by a negative number, the inequality sign must be flipped,
  • therefore, the correct answer is
x < − 3
Solve − 6x − 8 > 40
  • Add 8 to both sides
  • − 6x > 48
  • Divide both sides by − 6
  • x > − 8
  • Recall that whenever you divide inequalities by a negative number, the inequality sign must be flipped,
  • therefore, the correct answer is
x < − 8
Solve − 9x + 14 > − 40
  • Subtract 14 from both sides
  • − 9x > − 54
  • Divide both sides by − 9
  • x > 6
  • Recall that whenever you divide inequalities by a negative number, the inequality sign must be flipped,
  • therefore, the correct answer is
x < 6
Solve [( − 2(x − 2))/( − 4)] + 2(2 − 3x) ≤ 6 − x
  • 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
x − [2/3]
Solve [(3(x − 4))/( − 5)] + 3(2 − 4x) ≤ 4 − 2x
  • 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
x[22/53]
Solve [(3(x + 4))/( − 2)] + 2(2 − 5x)6 + 5x
  • 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
x ≤ − [16/33]
Solve [(5(x − 3))/( − 3)] − 4(3 − 5x)10 + 6x
  • 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
x[51/37]

*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

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 inequality0161

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