5 l 3x-2 l = 10

First let us divide by 5:

==> 5 l 3x-2 l /5 = 10/5

==> l 3x -2 l = 2

Now we have two cases:

case (1):

3x-2 = 2

Add 2 to both sides:

==> 3x = 4

Now divide by 3:

==> x= 4/3

Case (2):

-(3x-2) = 2

==> -3x + 2 = 2

Subtract 2 from both sides:

==> -3x =0

Now let us divide by -3:

==> x= 0

**Then there are two possible values for x: **

**x = { 0, 4/3}**

We'll solve the equation, expressing first the modulus.

Case 1)

l 3x-2 l = 3x - 2 for 3x-2 >= 0

3x =< 2

x =< 2/3

Now, we'll solve the equation:

5(3x-2) = 10

3x-2 = 2

3x = 4

x = 4/3

Since x = 4/3 does belong to the interval of admissible values,[2/3, +infinite], we'll accept it.

Case 2)

l 3x-2 l = -3x + 2 for 3x-2 < 0

3x<2

x<2/3

Now, we'll solve the equation:

5(-3x+2) = 10

-3x+2 = 2

-3x = 0

x = 0

Since x = 0 does belong to the interval of admissible values, (-infinite, 2/3), we'll accept it.

To solve the equation 5|3x-2| = 10.

We divide both sides of the given equation by 5:

|3x-2| = 10/5 = 2.

There are two cases: 3x-2 > 0 and 3x-2 < 0.

Case (1): 3x-2 > 0.

When 3x-2 > 0, |3x-2| = 3x-2.

So 3x-2 = 2.

3x = 2+2 = 4.

x = 4/3.

Case (2): when 3x-2 < 0.

When 3x-2 < 0, |3x-2| = -(3x-2) = 2-3x.

Therefor 2-3x = 2.

2-2 = 3x

0 = 3x. Or

x = 0.

So the solution is x = 0 or x = 4/3.

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