MHB Absolute Value Equation |3x - 2|/|2x - 3| = 2

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To solve the absolute value equation |3x - 2|/|2x - 3| = 2, the first step is to set |3x - 2| equal to 2|2x - 3|. This leads to two cases: 3x - 2 = 4x - 6, resulting in x = 4, and 3x - 2 = -4x + 6, which gives x = 8/7. The discussion emphasizes the importance of not dropping the absolute value signs and correctly squaring both sides when necessary. It is clarified that when squaring, the constant must also be squared, which is crucial for obtaining the correct solutions. The final quadratic equation derived does not factor easily, indicating a potential sign error in earlier calculations.
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Solve the absolute value equation.

|3x - 2|/|2x - 3| = 2

Solution:

|3x - 2| = 2|2x - 3|

3x - 2 = 2(2x - 3)

3x - 2 = 4x - 6

Solving for x, I get x = 4.

However, the textbook has two answers for this problem.
The answer is also 8/7.

How do I find 8/7?
 
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RTCNTC said:


Solution:

|3x - 2| = 2|2x - 3|

3x - 2 = 2(2x - 3)

As MarkFL pointed out in https://mathhelpboards.com/pre-algebra-algebra-2/absolute-value-equation-1-a-25152.html, you can’t just drop the absolute-value signs just like that! Otherwise you could have
$$|1|=|-1|\ \implies\ 1=-1.$$
The other solution you missed was
$$3x-2\ =\ -2(2x-3).$$
Alternatively, you can also do what you did in https://mathhelpboards.com/pre-algebra-algebra-2/absolute-value-equation-2-a-25153.html: square both sides.
 
3x−2 = −2(2x−3)

3x - 2 = -4x + 6

3x + 4x = 6 + 2

7x = 8

x = 8/7

- - - Updated - - -

Olinguito said:
As MarkFL pointed out in https://mathhelpboards.com/pre-algebra-algebra-2/absolute-value-equation-1-a-25152.html, you can’t just drop the absolute-value signs just like that! Otherwise you could have
$$|1|=|-1|\ \implies\ 1=-1.$$
The other solution you missed was
$$3x-2\ =\ -2(2x-3).$$
Alternatively, you can also do what you did in https://mathhelpboards.com/pre-algebra-algebra-2/absolute-value-equation-2-a-25153.html: square both sides.


If I decide to square both sides, must I also square 2?

Like this:

|3x - 2|^2 = [2|2x - 3|]^2

or

Like this:

|3x - 2|^2 = 2[|2x - 3|]^2
 
RTCNTC said:
3x−2 = −2(2x−3)

3x - 2 = -4x + 6

3x + 4x = 6 + 2

7x = 8

x = 8/7

- - - Updated - - -
If I decide to square both sides, must I also square 2?

Like this:

|3x - 2|^2 = [2|2x - 3|]^2

or

Like this:

|3x - 2|^2 = 2[|2x - 3|]^2
Yes, you have to square the 2 as well. [math](a (x - 1))^2 = a^2 (x - 1)^2[/math] for example.

-Dan
 
topsquark said:
Yes, you have to square the 2 as well. [math](a (x - 1))^2 = a^2 (x - 1)^2[/math] for example.

-Dan

It really helps to know that 2 must also be squared.
 
Take a look at my reply. The final quadratic equation does not factor leading to the textbook answers.

View attachment 8536
 

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You didn't square the 2 on the RHS.
 
MarkFL said:
You didn't square the 2 on the RHS.

You are right.
 
I squared 2 on the right side but ended up with a quadratic equation that does not lead to the textbook answers. See picture.

View attachment 8537
 

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  • #10
You've made a sign error, you should get:

$$7x^2-36x+32=0$$
 
  • #11

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