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Introduction to Absolute Value

Introduction to Absolute Value

Assessment

Presentation

Mathematics

9th - 12th Grade

Practice Problem

Medium

CCSS
6.NS.C.7C, 6.EE.B.5

Standards-aligned

Created by

Natalie McIntosh

Used 39+ times

FREE Resource

10 Slides • 13 Questions

1

Introduction to Solving Absolute Value Equations

Let's start at the beginning!

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2

Multiple Choice

The absolute value of a number represents the ___________ from that number to zero on a number line.

1

distance

2

journey

3

transformation

4

dissolution

3

Multiple Choice

Evaluate: |-6|

1

6

2

-6

3

1

4

0

4

Multiple Choice

Evaluate: |6|

1

6

2

-6

3

0

4

1

5

Absolute Value

The distance from zero on a number line.

| -6 | = 6 means that the value -6 is six steps away from zero on a number line.


So what if we know the distance? How can we figure out what the value is?

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6

Multiple Select

Which of the following could I substitute in for x and get 3?

 x=3\left|x\right|=3  

1

3

2

-3

3

0

4

1

7

There are 2 answers!

"What are two numbers that are 3 units away from 0?"

Two values work!

x = -3 or x = 3

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8

Multiple Select

What about this one? Which of the following values could I substitute in for x and it would make a true sentence?

 x2=3\left|x-2\right|=3  

1

3

2

-3

3

-1

4

5

5

2

9

Use Substitution

Let x = -1

| -1 - 2 | = | -3 | = 3

Let x = 5

| 5 - 2 | = | 3 | = 3

Slide image

10

What about this one?

|2x+7| = 5

If we weren't given multiple choice answers, we could guess until we found values that worked.

However, there are steps to solve.

11

Steps for Solving Absolute Value Equations

  • ISOLATE the Absolute Value BARS

  • SPLIT into two equations: the positive case and the negative case

  • SOLVE each equation

  • CHECK each solution using substitution into the original equation

12

Multiple Choice

What does it mean to "isolate the bars?"

How would you isolate the bars for this equation:

 x3 = 10\left|x\right|-3\ =\ 10  

1

Add 3 to both sides

2

Subtract 3 from both sides

3

Add 10 to both sides

4

Subtract 10 from both sides

13

We isolate the bars just as we would isolate a variable in a regular equation.

 x3=10\left|x\right|-3=10  
 x3+3 = 10+3\left|x\right|-3+3\ =\ 10+3  (Adding 3 to both sides)
 x=13\left|x\right|=13  


For this problem, we see a couple of solutions for x.

-13 and 13

14

Multiple Choice

How about this one? How would you isolate the bars?

 4x=164\left|x\right|=16  

1

Add 4 to both sides

2

Subtract 4 from both sides

3

Multiply 4 on both sides

4

Divide both sides by 4

15

Multiple Choice

How about this one?
Isolate the bars:

 x4=16\frac{\left|x\right|}{4}=16  

1

Add 4 to both sides

2

Subtract 4 from both sides

3

Multiply 4 on both sides

4

Divide both sides by 4

16

Multiple Choice

How about this one?
Isolate the bars:

 4x+2=164\left|x+2\right|=16  

1

Add 2 to both sides

2

Subtract 2 from both sides

3

Multiply 4 on both sides

4

Divide both sides by 4

17

Whatever is inside the bars...

...stays inside the bars when you are isolating. Don't mess with the inside!

18

Multiple Choice

Isolate the bars......

 x+3+2=8\left|x+3\right|+2=8  

1

 x+3=6\left|x+3\right|=6  

2

 x=3\left|x\right|=3  

3

 x+5=8\left|x+5\right|=8  

4

 5x=8\left|5x\right|=8  

19

Multiple Choice

Isolate the bars......

 5x87=135\left|x-8\right|-7=13  

1

 x8=4\left|x-8\right|=4  

2

 x=12\left|x\right|=12  

3

 5x40=20\left|5x-40\right|=20  

4

 5x47=13\left|5x-47\right|=13  

20

So we know how to isolate the bars...then what?

Split into two equations

21

Split into two equations: 

 x+3=6\left|x+3\right|=6  

  • The positive case:  x+3=6x+3=6  which can be solved --> x = 3

  • The negative case:  x+3=6x+3=-6  which can be solved --> x = -9

22

Multiple Select

Split into two equations:

 2x+7=5\left|2x+7\right|=5  

1

 2x+7=52x+7=5  

2

 2x+7=5-2x+7=5  

3

 2x+7=52x+7=-5  

4

 2x7=5-2x-7=-5  

23

Multiple Select

Isolate the bars and Split into two equations:

 x8+2=5\left|x-8\right|+2=5  

1

 x8=3x-8=3  

2

 x8+2=5x-8+2=5  

3

 x8=3x-8=-3  

4

 x8+2=5x-8+2=-5  

Introduction to Solving Absolute Value Equations

Let's start at the beginning!

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