You need to use absolute value definition, such that:

`|x| = {(x, x>=0),(-x, x< 0):}`

Reasoning by analogy yields:

`|2x+1| = {(2x+1, 2x+1>=0),(-2x-1, 2x+1< 0):} `

`|2x+1| = {(2x+1, x>=-1/2),(-2x-1, x< -1/2):}`

You need to solve the equations, such that:

`{(2x + 1 = 52/26,x>=-1/2),(2x + 1 = -52/26, x < -1/2):}` => `{(2x = 2 - 1, x in [-1/2,oo]),(2x = -2 - 1, x in (-oo,-1/2)):}`

`{(x = 1/2 = 0.5 in [-1/2,oo]),(x = -3/2 = -1.5 in (-oo,-1/2)):}`

**Hence, evaluating the solutions to absolute value equation yields **`x = -1.5, x = 0.5.`

First we'll divide both sides by two:

13*|2x+1|=26

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

Case 1:

l 2x + 1 l = 2x + 1 for 2x + 1 >= 0

2x >= -1

x >= -1/2

Now, we'll solve the equation:

13(2x + 1) = 26

We'll divide by 13:

2x + 1 = 2

2x = 2-1

**x = 1/2**

Since x =1/2 is in the interval of admissible values,[-1/2, +infinite], we'll accept it.

Case 2:

l 2x + 1l = -2x - 1 for 2x + 1 < 0

2x < -1

x < -1/2

Now, we'll solve the equation:

13(-2x - 1) = 26

-2x - 1 = 2

-2x = 3

**x = -3/2**

**Since x = -3/2 is in the interval of admissible values, (-infinite, -1/2), we'll accept it.**