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INSTRUCTORSCarleen EatonGrant Fraser
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Lecture Comments (14)

1 answer

Last reply by: Dr Carleen Eaton
Wed Jan 1, 2014 12:44 AM

Post by Myriam Bouhenguel on December 22, 2013

In the example you gave in the beginning about the 3 |5x+4| = 6, instead of dividing the 3 to both sides can I also distribute the 3 with the 5x+4 inside the absolute value and solve the rest of the equation?

0 answers

Post by julius mogyorossy on July 18, 2013

Did Dr. Carleen get the wrong answer for example one, it seems so to me, but I am defective, but soon I shall be perfect, can't wait, today really proved the truth of it.

2 answers

Last reply by: Manfred Berger
Tue May 28, 2013 6:17 AM

Post by Su Jung Leem on August 7, 2012

Dr. Eaton, I have the same question as the girl above. If you actually solved the equation, how can there be invalid result in the last example? I'm a bit confused

1 answer

Last reply by: Dr Carleen Eaton
Tue Jul 10, 2012 11:02 PM

Post by Sayaka Carpenter on July 9, 2012

Dr. Eaton, if you get the answer by solving the equation, how could it not be valid?

1 answer

Last reply by: Dr Carleen Eaton
Thu Feb 23, 2012 11:35 AM

Post by julius mogyorossy on February 13, 2012

Dr. Eaton, I could not understand Absolute Value Equations when Grant talked about them, but you made me understand them, thanks. I was troubled by some things you did in Ex.1, but maybe in time I shall understand them. Thanks.

3 answers

Last reply by: Venugopal Ghanta
Mon Jun 27, 2011 10:26 AM

Post by Jonathan Bergan on July 29, 2010

Dr. Carleen Eaton, could you further explain example IV in regards to why you multiplied by -1? Thanks

Solving Absolute Value Equations

  • The absolute value of a number is its distance from 0.
  • The absolute value of x is equal to x if x is greater than or equal to 0; otherwise, it is equal to –x.
  • Use this definition of absolute value to solve equations involving absolute values. Split the original equation |x| = a into two different possibilities, or cases, based on the definition: x = a and x = −a. Solve each one and combine the solutions.
  • Some equations have no solutions.
  • Always check all of the solutions. Some may not satisfy the original equation and must be discarded.

Solving Absolute Value Equations

Evaluate for x = − 5 |2x − 1| + |x − 2| − 2| − x|
  • Plug in x = − 5 into the expression
  • |2( − 5) − 1| + |( − 5) − 2| − 2| − ( − 5)|
  • Simplify
  • | − 11| + | − 7| − 2|5|
  • 11 + 7 − 2(5)
  • 11 + 7 − 10
= 8
Evaluate for x = 2 | − x| + 2|x − 5| − |2 − x|
  • Plug - in x = 2 in to the expression
  • | − (2)| + 2|(2) − 5| − |2 − (2)|
  • Simplify
  • | − 2| + 2| − 3| − |0|
  • 2 + 2(3)
= 8
Solve: 2|4x − 2| = 36
  • Divide both sides by 2
  • |4x − 2| = 18
  • Break the absolute value into two equations and solve for x
  • 4x - 2 = 18
    4x = 20 4x=-16
    x = 5
  • 4x -2 = -18
    4x = -16
    x = -4
x = 5 and x = -4
Solve: − 3|5x − 3| = − 9
  • Divide both sides by − 3
  • |5x − 3| = 3
  • Break the absolute value into two equations and solve for x
  • 5x − 3 = 3
    5x = 6
    x = [6/5]
  • 5x − 3 = − 3
    5x = 0
    x = 0
x = [6/5] and x = 0
Solve: − 2| − 9b| = − 54
  • Divide both sides by − 2
  • | − 9b| = 27
  • Break the absolute value into two equations and solve for x
  • − 9b = 27
    − 9b = 27 b = −3
  • − 9b = − 27
    −9b = −27
    b = 3
b = −3 and b = 3
Solve: 2 − 3|2n − 7| = − 16
  • Subtract 2 from both sides
  • − 3|2n − 7| = − 18
  • Divide both sides by − 3
  • |2n − 7| = 6
  • Break the absolute value into two equations and solve for x
  • 2n − 7 = 6
    2n = 13
    n = [13/2]
  • 2n − 7 = − 6
    2n = 1
    n = [1/2]
n = [13/2] and n = [1/2]
Solve:
10| − 8 + 2a| + 10 = 30
  • Subtract 10 from both sides
  • 10| − 8 + 2a| = 20
  • Divide both sides by 10
  • | − 8 + 2a| = 2
  • Break the absolute value into two equations and solve for x
  • − 8 + 2a = 2
    2a = 10
    a = 5
  • − 8 + 2a = − 2
    2a = 6
    a = 3
a = 5 and a = 3
Solve: |4x − 5| = − 10
  • No Solution.
Recall that the absolute value represents the distance between a number and zero. Thefore, distance cannot be negative, for this reason, this problem has no solution.
Solve: − 5|6x + 10| = 20
  • Divide both sides by − 5
  • |6x + 10| = − 4
  • No Solution.
Once again, the distance between a number and zero is always positivie. Therefore, there is no solution.
Solve: − 5|6x + 10| − 30 = − 20
  • Add 30 to both sides
  • − 5|6x + 10| = 10
  • Divide both sides by − 5
  • |6x + 10| = − 2
  • No solution.
The distance between a number and zero is always positive. A negative solution means there is no answer to the absolute value equation.

*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 Absolute Value Equations

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
  • Absolute Value Expressions 0:09
    • Distance from Zero
    • Example: Absolute Value Expression
  • Absolute Value Equations 1:50
    • Example: Absolute Value Equation
    • Example: Isolate Expression
  • No Solution 3:46
    • Empty Set
    • Example: No Solution
  • Number of Solutions 4:46
    • Check Each Solution
    • Example: Two Solutions
    • Example: No Solution
    • Example: One Solution
  • Example 1: Evaluate for X 7:16
  • Example 2: Write Verbal Expression 9:08
  • Example 3: Solve the Equation 12:18
  • Example 4: Simplify Using Properties 13:36

Transcription: Solving Absolute Value Equations

Welcome to Educator.com.0000

In today's lesson, we are going to be covering solving absolute value equations.0002

Recall that the absolute value x, written as |x| (this is the symbol for absolute value), is the distance from x to 0 on a number line.0009

For example, if you have the absolute value of x equals three, what you are saying is0022

that the absolute value of this number is 3 away from 0 on the number line (1, 2, 3).0036

OK, so looking at this: if I go from 0 to 3, this is 3 units away from 0 on the number line; therefore, x could equal 3,0052

because 3 is 3 away from 0 on the number line.0070

However, consider this: -3 is also 3 units away from 0 on the number line.0074

So, x could also equal -3; so, if the absolute value of x equals 3, x could be 3 (since the absolute value of 3 is 3),0081

and x could be -3 (since the absolute value of -3 is also 3).0093

Knowing this, and really understanding this definition, will allow you to solve equations involving absolute value.0102

Again, absolute value equations: you can solve equations containing absolute values, using the definition of absolute value.0110

For example, |9x + 2| = 29; well, I already said that, if the absolute value of x is 3, that means that x equals 3, or x equals -3.0121

So, you can apply that same concept right here: 9x + 2 = 29, or 9x + 2 = -29.0134

So, once you remove the absolute value bars (and you can do that by turning it into two related equations,0144

one where it equals the positive, and one where it equals the negative value)--once you have done that,0151

all you need to do is solve each equation.0155

So, I am going to solve, just using my usual techniques: subtract two from both sides, and that gives me 9x = 27.0159

Over here, if I subtract 2 from both sides, I am going to get 9x = -31.0169

Then, I am going to divide both sides by 9.0176

OK, so you handle the absolute value equations the same way, even if it is more complex.0186

The other thing to keep in mind is that sometimes, you have to first isolate the absolute value expression on the left side of the equation.0193

For example, if you were given 3 times |5x + 4| equals 6, the first step would be to divide both sides by 3,0199

because once you have done that, then you have the absolute value isolated,0210

and you can proceed by saying 5x + 4 = 2 or 5x + 4 = -2, and then solving both of those.0214

In some situations, some absolute value equations, where the absolute value equals c, if c is less than 0, these have no solution.0227

So, if it says that the absolute value of x is a negative number, then there is no solution; and we just say that the solution is the empty set.0236

And to make this more concrete: if I said that the absolute value of x equals -4...well, there is no situation0249

where an absolute value is going to be a negative number, because remember: the absolute value is defined0258

as the distance between that absolute value and 0 on the number line.0266

And you can't have negative distance; so there is no solution here.0271

And instead, we either just write the empty set as such, or like this, indicating that there is no solution.0275

We just talked about a situation where there is no solution.0287

An absolute value equation can have 0 solutions (which we just discussed--it is the empty set), 1 solution, or 2 solutions.0290

And it is important to check each answer to make sure that it is a valid solution.0298

For example, if I have |x + 4| = 9, I am going to go ahead and solve that,0304

using the usual technique of turning it into two related equations, x + 4 = 9 and x + 4 = -9.0311

I am going to solve that: x = 5; here, I am going to get x = -13.0321

So, let's check this by going back to the original, |x + 4| = 9.0329

If x = 5, then I am going to get |5 + 4| = 9; so, the absolute value of 9 equals 9--and that is true.0335

Since this is true, this is a valid solution; x = 5 is a valid solution.0345

OK, doing the same thing for my other solution: |x + 4| = 9...0350

trying this |-13 + 4| = 9, well, -13 + 4 is -9, so the absolute value of -9 is 9.0358

This is also a valid solution, since this is a true statement.0368

So here, I have two solutions: previously, we discussed that, if you end up with something like |x| = -6, there is no solution; it is the empty set.0372

The situation where you can get one solution is if you end up with |x| = 0.0390

So, if I went to solve this, I would say, "x equals +0, and x equals -0"; but that doesn't really just make sense; they are just 0.0400

Therefore, there is only one solution: x = 0; so here, I have one solution.0408

Three possibilities: no solution, one solution, or two solutions.0414

Or you may think you have two solutions, and then you go back and plug them in, and find out one is not valid.0419

And in that case, you could have something like this, that appears that it would end up with two solutions,0425

and it only ends up with one, or possibly even none.0430

OK, Example 1: Evaluate for x = -3; and this is multiple absolute value terms here--0437

substituting in -3 for x, solving for the absolute value of this would be 3 times -3...that is going to give me |-9 - 4|;0447

here, I have 2 times -3; that is |-6 + 3|; minus 3 times...the absolute value of a negative, times a negative, is a positive.0469

OK, this is |-9 - 4|; that is the absolute value of -13, plus |-6 + 3|; that is -3; minus 3...absolute value of 3.0480

Now, I just need to find the absolute value for each of these.0493

Well, the absolute value of -13 is 13; the absolute value of -3 is 3; the absolute value of 3 is 3;0497

but this time, we are multiplying it times a -3; this is -3 times |3| (which is 3).0512

So here, it's 13 + 3...-3 times 3 is -9, so 13 + 3 is 16, minus 9 is 7.0523

OK, so we are solving this by substituting in -3 for x, finding the absolute values for these three, and then this one is multiplied by -3, and then simply adding.0536

Example 2: I have an absolute value expression on the left, but it is not isolated.0550

So, the first step is to isolate the absolute value, and then find the two related equations and solve them.0555

Divide both sides by 4 to get that isolated.0562

Now, recall that the absolute value of x equals 2, for example: this means that x could equal 2,0570

or x could equal -2--to just illustrate the definition of absolute value.0578

In order to get rid of these absolute value symbols, I am going to create two related equations,0585

3x + 4 = 12, or 3x + 4 = -12: then I am going to solve both, and check to make sure that they are valid solutions.0590

3x = 8; so divide both sides...I had 3x + 4, so I subtracted 4 from both sides.0602

And then, over here, I have 3x = -16, subtracting 4 from both sides--I am just doing these together.0619

Now, dividing both sides by 3 is going to give me x = 8/3; dividing here, I get -16/3.0631

Now, check each of these in the original: that is this 4 times the absolute value of 3x + 4 equals 48.0641

First checking this one: 4, and this is 3 times 8/3 + 4, equals 48...so the 3's cancel out, and that gives me 8 + 4...equals 48,0652

so 4 times the absolute value of 12 equals 48; the absolute value of 12 is 12, so 4 times 12 equals 48; and it does, so this is valid.0672

This first solution is valid.0685

Now, substituting in -16/3, again, in this original: that is 4 times |3(-16/3) +4| = 48.0688

The 3's cancel out; that gives me 4|16 + 4| = 48.0704

-16 +4 is -12...equals 48...the absolute value of -12 is 12; so again, I come up with 4 times 12 equals 48, and that is valid.0710

That is a true statement; so both of these solutions are valid.0724

So this time, I had two solutions, 8/3 and -16/3, that satisfy this absolute value equation.0729

Example 3: again, my first step is to isolate the absolute value expression on the left side of the equation.0740

And I am going so start out by subtracting 15 from both sides, and that is going to give me -3 on the right.0748

Now, I want to divide both sides by 3, and that is going to give me |2x - 4| = -1.0756

And I don't even need to go any farther, because what this is saying is that0766

the absolute value of whatever this expression ends up being (once I solve for x)--the absolute value of this equals -1.0769

Well, that is not valid; you cannot have an absolute value equal a negative number, because it violates the definition of absolute value.0778

Therefore, I don't even need to go any farther; I can just say that there is no solution, or that it is the empty set.0789

So, there is no solution to this absolute value equation.0795

The important thing is to just look carefully at your work and make sure you didn't make any mistakes.0802

And if you did all the math correctly and handled this correctly, and you come up with something like this, then you didn't do anything wrong.0806

It is just that there is no solution.0813

OK, another absolute value expression: this time, the absolute value is already isolated on the left.0817

So, looking at this, it looks more complex; but we just use the same logic that we did with the simple case.0824

If the absolute value of x is 2, x equals 2, or x equals -2.0834

So, I do the same thing here: I get rid of the absolute value bars, and this is my positive permutation.0838

And then, I also have 2x - 7 = -(3x + 8).0846

OK, solving each of these: I am going to add 7 to both sides; that is going to give me 2x = 3x +15.0854

Then, I am going to subtract 3x from both sides, which is going to give me -x = 15.0867

I am going to multiply both sides by -1 to get x = -15.0876

So, that is my first solution: solving this...this is 2x - 7 = -3x - 8.0884

Adding 7 to both sides is 2x = -3x -1.0893

Adding 3x to both sides: 5x = -1; divide both sides by 5: x = -1/5.0900

OK, check: with absolute value equations, you always have to check your solutions.0909

So, checking this back in the original equation: 2 times -15, minus 7--the absolute value of that--equals 3 times -15, plus 8.0917

OK, so I have 2(-15), which is -30, minus 7, equals 3(-15), which is -45, plus 8.0931

This gives me |-37| equals...well, -45 + 8 is -37; the absolute value of -37 is 37.0945

Well, 37 does not equal -37, so this is not a valid solution; this is not true--this did not satisfy this equation.0959

So, x = -15 is not a valid solution; let's try this one, x = -1/5, substituting it in here.0970

|2(-1/5) - 7| = 3(-1/5) + 8: that is going to give me |(-2/5)-7| = (-3/5) + 8.0977

So then, adding these two together, I am going to get |-7 2/5| = 8 - 3/5...is 7 2/5.0999

OK, the absolute value of -7 2/5 is 7 2/5, equals 7 2/5; and that is true; this is a valid solution.1010

So, I actually have only one solution to this equation, and it is x = -1/5.1022

We are starting out by breaking this into two related equations and removing the absolute bar,1030

solving each, getting two solutions, and then checking and finding out that the first one is not valid, and that the second one is valid.1036

That concludes this session for Educator.com, and I will see you for the next Algebra II lesson.1046