Sign In | Subscribe
INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Algebra 1
  • Discussion

  • Study Guides

  • Practice Questions

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Bookmark and Share
Lecture Comments (5)

2 answers

Last reply by: Denise Bermudez
Sun Mar 8, 2015 10:06 AM

Post by Denise Bermudez on March 7, 2015

hi
I dont understand why in example II when you break the absolute values into two separate ones (the second one)you get 5x + 2? Shouldnt you get a 5x-2 when you multiply it by a (-1)?
because in example III when you did (x+2)=-1 and multiplied it you got
-x-2=-1

1 answer

Last reply by: Professor Eric Smith
Mon Dec 2, 2013 8:42 PM

Post by Jason Arias on November 26, 2013

Does absolute value "||" and parenthesis "()" have the same priority under the order of operations? Watching the order in which you solve the solutions implies this, but i just want to confirm.

Apart from "||" changing whatever is inside into a positive, seems to behave like "()"

Solving Equations with Absolute Values

  • When working with an expression inside an inequality, we must account for the expression being either positive or negative during the solving process. We do this by isolating the absolute value, then splitting the problem into two.
  • It is a good idea to check your work on these problems by substituting the solution back in the original.

Solving Equations with Absolute Values

| 4x − 10 | = 12
  • 4x − 10 = 124x − 10 = − 12
  • 4x − 10 = 12
  • 4x = 22
  • x = [22/4]
  • x = 4[2/4] = 4[1/2]
  • 4x − 10 = − 12
  • 4x = − 2
  • x = − [2/4]
x = − [1/2]
| 6x − 13 | = 11
  • 6x − 13 = 116x − 13 = − 11
  • 6x − 13 = 116x = 24x = 4
  • 6x − 13 = − 116x = 2x = [1/3]
x = 4 or x = [1/3]
| 8k + 6 | = 24
  • 8k + 6 = 248k + 6 = − 24
  • 8k + 6 = 24
  • 8k = 18
  • k = [18/8] = [9/4] = 2[1/4]
  • 8k + 6 = − 24
  • 8k = − 30
  • k = − [30/8] = − [15/4] = − 3[3/4]
k = 2[1/4] or − 3[3/4]
| − 4t − 10 | = 6
  • − 4t − 10 = 6− 4t − 10 = − 6
  • − 4t − 10 = 6
  • − 4t = 16
  • t = − 4
  • − 4t − 10 = − 6
  • − 4t = 4
  • t = − 1
t = − 4 or t = − 1
| − 2g − 8 | = 22
  • − 2g − 8 = 22− 2g − 8 = − 22
  • − 2g = 30
  • g = − 15
  • − 2g − 8 = − 22
  • − 2g = − 14
g = 7
| − 3j + 15 | = 9
  • − 3j + 15 = 9− 3j + 15 = − 9
  • − 3j = − 6
  • j = 2
  • − 3j + 15 = − 9
  • − 3j = − 24
  • j = 8
j = 2 or j = 8
| − v + 5 | = 6
  • − v + 5 = 6− v + 5 = − 6
  • − v + 5 = 6
  • − v = 1
  • v = − 1
  • − v + 5 = − 6
  • − v = − 11
  • v = 11
v = − 1 or v = 11
Graph the function
y = |x| − 3
  • Make a table of points
  • x
    line
    0
    − 1
    1
    − 3
    3
    − 5
    5
    y
    line
    − 3
    − 2
    − 2
    0
    0
    2
    2
  • Graph utilizing the coordinate points
Graph the function
y = | − x + 3| + 1
  • Make a table of points
  • x
    line
    0
    1
    3
    5
    7
    9
    11
    y
    line
    4
    3
    1
    3
    5
    7
    9
  • Graph utilizing the coordinate points
Graph the function
y = − |[x/2] + 1|
  • Make a table of points
  • x
    line
    0
    2
    − 2
    4
    − 4
    − 6
    6
    y
    line
    − 1
    − 2
    0
    − 3
    1
    2
    − 4
  • Graph utilizing the coordinate points

*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 Equations with Absolute Values

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
  • Objectives 0:08
  • Solve Equations with Absolute Values 0:18
  • Solve Equations with Absolute Values Cont. 1:11
    • Steps to Solving Equations with Absolute Values
  • Example 1 3:23
  • Example 2 6:34
  • Example 3 10:12

Transcription: Solving Equations with Absolute Values

Welcome back to www.educator.com.0000

In this lesson we are going to go ahead and look at solving equations that have absolute values in them.0002

We only have one goal, how do you get down to the nuts and bolts of solving it when you see an absolute value.0010

When you have a number that is equal to an absolute value or any algebraic expression, then you can break it down into the following.0021

That expression is equal to the number or that expression is equal to the negative of that number.0030

What this is trying to say is that when you are dealing with this situation, absolute value equals number, is that you can break it down into 2 situations.0035

It takes care of 2 possibilities that whatever was inside the absolute value could have been positive 0045

or maybe what is on the inside could have been negative.0050

The reason why we have 2 possibilities is the absolute values going to take away information about what the original sign was.0053

Maybe that original sign was positive, maybe it was negative, but we are not sure so we have to take them both into account.0060

To highlight how this works, let us look at some actual numbers.0067

Let us suppose that our algebraic expression was just x and we are going to look at the absolute value of x is equal to 2.0073

What would have to be inside that absolute value in order for this thing to work? 0081

One possibility is that maybe x = 2, so we get this case over here.0085

What if the x was negative and we lost all information about that negative sign.0091

That looks like more of like the situation on the right, what we see is that x could be 2 or x could actually be -2.0097

The only difference between these two is I multiply both sides by -1.0105

What I like to do when handling these absolute values is immediately split them into 2 different problems.0111

I will handle what I called my positive case where I just keep everything the same and drop the absolute values.0118

Then I will take care of what I call the negative case, 0125

I will put a negative sign on whatever was inside the absolute value and then continue solving from there.0127

When I’m all done, I will go ahead and take both of my solutions and check them in the original problem, just to make sure that they are correct.0133

Some other things to consider when solving an equation that have an absolute value is to make it fit the form.0143

You want to make sure that you isolate the entire absolute value first.0150

Do not worry about splitting it into 2 problems, until you have absolute value on one side and the one on all the numbers and other stuff on the other. 0155

Once you do have that absolute value, go ahead and split it into 2 problems.0165

And what you are looking to do with each of those problems is you want one of them to handle what I said positive possibilities,0172

it will be exactly the same, just no absolute value.0180

And one to handle the negative possibilities, what would happen if there was a negative sign on the inside of those absolute values.0183

When you have two new problems then simply solve each of them separately as normal. 0191

Follow all of your normal rules for algebra and you should be okay. 0196

Let us give this a try. 0201

The first one I'm looking at the absolute value x -3 -1 = 4. 0204

The very first thing I want to do is to find those absolute values here and get them all alone onto one side of my equation. 0211

I have the absolute value of x - 3 and isolate it then I will add one to both sides.0220

Once I get the absolute value completely isolated, this is the spot we are going to split this into 2 problems. 0230

Get some space and get our two problems.0237

On the left side here, I'm going to consider this my positive possibility that whatever the absolute value was encapsulating it was positive.0241

It is one of the changes that I need to make.0250

The other possibility is that maybe the x -3 maybe that entire thing was negative.0254

Notice how I put a negative sign in there.0258

Now that we have two problems, we just go through each one separately and see how they turn out.0262

x - 3 = 5, add 3 to both sides and get that x = 8.0268

With the other one, it is okay if you want to distribute that negative sign first.0276

-x + 3 = 5, then we can subtract 3 and move that negative sign, x = -2.0288

I have two solutions to my equation that I started with.0300

It is a good idea to check these just to make sure they work out.0306

Let us take them individually and see what happens. 0309

8 - 3 would give us 5.0319

Our absolute values make everything positive so the absolute value of 5 would be 5.0326

I can see that sure enough, 5 – 1 =4 and that solution checks out.0333

Onto the other side, we will put a -2 – 3 working with the inside of the absolute values,-2 - 3 would give us -5.0340

Notice, as we start to simplify that we can see what I was talking about how the sign got me positive so we do have to handle both possibilities.0358

Cancel the absolute value of -5 is 5, 5 - 1 = 4 and sure enough, we know that is true so our solution checks out, that x = 8 or x = -2.0373

In this next example, we will have to do a little bit more on getting the absolute value isolated, but again you will see this is not too bad.0397

To get rid of that fraction out front, I'm going to multiply both sides by 2, 2 over there and 2 over there.0404

It will give us the absolute value of 5x + 2 = 12.0415

We have the absolute value on one side of the equal sign.0422

Let us go ahead and split this into two different problems. 0427

Here I will handle my positive possibility that whatever was inside absolute value it is positive.0435

Over here, maybe what was inside the absolute values was negative, so we will put that negative sign.0443

Let us solve both of these separately.0453

On the left, I will subtract 2 from both sides, and then we will go ahead and divide both sides by 5, I have x = 2.0457

On the other side, I’m going to do something a little bit different.0471

To move that negative sign at the very beginning, I will multiply both sides by -1. 0474

It is important to know that you can also distribute through being your first step and then go through the solving process.0481

I’m just doing this so I can take care of the negative sign at the very beginning.0487

Negative × negative on the left side would give me 5x + 2 =-12.0491

Let us subtract 2 from both sides, and finally divide by 5.0500

Now I have two solutions to my absolute value problem.0515

Let us take a little bit of time just to check them and make sure they work out.0521

½ of 5 × 2 + 2 does that equals 6?0527

Working on the inside of the absolute value, I have 5 × 2 = 10 + 2 =12.0536

The absolute value of 12 is 12, 1/2 of 12 = 6, that one checks out.0546

On to the other side, this one is -14/5 + 2.0556

Let us see if this one equal 6 or not.0570

14/5 × 5 = -14, -14 + 2 =-12, it is very similar to what we got last time.0573

We are just dealing with the negative on the inside.0593

The absolute value will make that -12 into a positive 12 and sure ½ of 12 = 6.0599

Both of our solutions work out just fine. 0607

There is still one more example to show you some of the things that might happen when going through the solving process.0615

This one is 3 times the absolute value of x + 2 + 3 = 0.0620

We will start off by trying to isolate that absolute value. 0626

I'm going to subtract 3 from both sides.0635

To get rid of that 3 out in front of the absolute value, let us divide both sides by 3.0639

I have the absolute value of x + 2 .0648

Now following our steps, let us split this into 2 problems.0651

Here we have the problem exactly the same, just no absolute value and over here, -x + 2 = -1.0660

Let us solve each of them separately.0670

I will subtract 2 from both sides given the x = -3. 0674

On the other side here, I will go ahead and distribute through with my negative sign and I will add 2 to both sides.0679

Lastly let us go ahead and multiply both sides by -1.0693

I have two possible solutions, x = -3 and x = -1.0699

Let us check them to see how they turn out.0705

Let us see, -3 + 2 = -1 then we will take the absolute value of -1, we get 1.0716

What I'm getting here is that 3 × 1 is 3 and when we add another 3, I get 6.0730

This shows that this one does not work out.0738

I'm going to get rid of that, it is not a solution.0742

Let us try it on the other side.0747

I will put in a -1.0750

I have 3, -1 + 2 = 1 and the absolute value of 1 is 1.0758

I have 3 × 1 + 3 I think that is the same situation as before or 6 = 0.0772

I think that does not work out as well so I have to throw away that solution. 0778

What this shows is that if none of the work then we actually have no solution.0784

Let us highlight that it is a good idea to always check your answers.0792

I will also highlight that some funny things could happen when working with absolute values.0796

Remember that an absolute value takes whatever you put into it and makes it positive.0800

In fact, we can see a problem in this step when we are working. 0806

I have that the absolute value of something is equal to a negative and that can not happen because our absolute value on one side will be positive.0813

On the other side we are setting equal to a negative and we can not get a positive and a negative, to agree.0826

That is why this one also has no solution.0836

If you catch it early on, it usually can save yourself a lot of work.0839

But if you do not, make sure your check your solutions also that you can definitely find these no solution problems.0842

Thank you for watching www.educator.com.0850