Unit 2: Topic 3: Equations Involving Absolute Values

Competencies

·   Apply the definition of absolute value

·   Solve equations involving absolute value

Suggested ways of teaching this topic: Teacher assisted jigsaw groups

Starter Activities

Obviously, the teacher shall start from the definition of absolute value. As students were familiar with it in the lower grades, the teacher shall ask students to state the definition of absolute values.

Expected Answer:   |x| = x if x>0, |x| = - x if x <0 and |x| = 0 ifx = 0.

Alternatively students might explain as:

When you take the absolute value of a number, you always end up with a positive number (or zero).  Whether the input was positive or negative (or zero), the output is always positive (or zero). For instance, | 3 | = 3, and | –3 | = 3 also.

If consensus is reached on the understanding of the definition the teacher may start with something simple like the following to be done in pairs or in small groups:

Example: Solve the following equations

a)      | x | = 6    

b)      | x + 2 | = 7    

c)      | x - 1 | = - 3     

d)     | 3 - x | = 5

Solutions:

a)       If you have x = –6, then " –x " indicates "the opposite of x", or, in this case, –(–6) = +6, a positive number. The minus sign in " –x " just indicates that you are changing the sign on x. It does not indicate a negative number comes out of the absolute value. This distinction can be crucial.

Thus, the solution is x = –6, 6

b)       Solve | x + 2 | = 7

To clear the absolute-value bars, we must split the equation into its possible two cases, one case for each sign:

(x + 2) = 7     or     –(x + 2) = 7
x + 2 = 7       or     –x – 2 = 7
x = 5             or     –9 = x

Thus, the solution is x = –9, 5.

c)      |x - 1 | = - 3      is meaningless! As from the definition an absolute value can’t be negative.

Thus, no solution is available.

d)     | 3 - x | = 5  here, we shall split the equation into its two possible two cases

(3 - x ) = 5 or –(3 - x ) = 5

3- x = 5  or – 3 + x = 5

x = -2 or x = 8

Thus, the solution is x = –2, 8.

Lesson Notes

The teacher can give equations of the following type to small groups where each group takes one question. At the end of the group work, let group leaders’ move to other groups to share their solutions and learn the others’ solutions.

Example: Solve the following equations

a)      | 2x+ 3 | = 2

b)      |  3– 2x  | = 1

c)      | 2(x – 3) | = 4

d)     | 2x – 3 | – 4 = 3

Solutions:

a)      | 2x + 3 | = 2 ; split the equation into its two possible cases

2x + 3 = 2 or –(2x + 3) = 2

2x + 3 = 2 or – 2x – 3 = 2

2x = –1  or –2x = 5

x =    or  

b)      | 3 – 2x | = 1; split the equation into its two possible cases

  3 – 2x = 1 or    – (3 – 2x) = 1

3 – 2x = 1 or    – 3 + 2x = 1

– 2x = – 2   or   2x = 4

x = 1  orx = 2

c)      | 2(x – 3) | = 4; split the equation into its two possible cases

2(x – 3)  = 4    or  –2(x– 3) = 4

2x –6  = 4    or  –2x+ 6  = 4

2x  =10  or      –2x  = –2

x  =5    or    x  = 1

 

d)     | 2x – 3 | – 4 = 3

First, we will isolate the absolute-value part; that is, we will get the absolute-value expression by itself on one side of the "equals" sign, with everything else on the other side:

| 2x – 3 | = 7

Now we will clear the absolute-value bars by splitting the equation into its two cases, one for each sign:

(2x – 3) = 7 or   –(2x – 3) = 7
2x – 3 = 7   or    –2x + 3 = 7
2x = 10       or    –2x = 4
x = 5           or    x = –2

Concluding Activities

The teacher shall make sure that the students got the idea of solving equations involving absolute value by asking them to tell him/her the general format of solving these types of equations. The students are expected to give the following summary:

Whatever the value of x might be, taking the absolute value of x makes it positive. Since x might have been positive and might have been negative, we have to acknowledge this fact when we take the absolute-value bars off, and you do this by splitting the equation into two cases. If the value of x was positive to start with, then you can bring that value out of the absolute-value bars without changing its sign. But x might also have been negative, in which case we would have to change the sign on x for the absolute value to come out positive.

To be able to remove the absolute-value bars, you have to isolate the absolute value onto one side, and then split the equation into the two possible cases. Then, start solving the two linear equations obtained from the two cases.

That is, |x| = x if x> 0, |x| = - x if x <0 and |x| = 0 if x = 0.

Practice Exercises

Solve the following equations

a)      | 4x – 1 | = 0                            

b)      | 2x+ 3 | = - 1   

c)      | 5x+ 3 | = 1

d)     | –5x – 1 |– 4 = 2

e)      | –2x+ 3 |– 2= – 1                   

f)       |– 3 –2x  | = 2

g)      |– 1 –5x  | + 4 =  2

h)      | –2(x+ 3) | + 2 = 3