# What is the value of x for the following equation: 26|2x+1|=52?

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We have to find x given that 26|2x+1|=52.

26|2x+1|=52

cancel 26 from both the sides.

=> |2x + 1| = 2

Now |x| represents the absolute value of x and has the same value for x as well as -x.

=> 2x + 1 = 2 and 2x + 1 = -2

=> 2x = 1 and 2x = -3

=> x = 1/2 and x = -3/2

**Therefore we get x is equal to 1/2 and -3/2**

To find x if 26|2x+1| = 52.

Solution:

26|2x+1| = 52.

We divide both sides by 26:

|2x+1| = 2.

|2x+1|=2 implies 2x+1 = 2, or 2x+1 = -2 = 2.

If 2x+1 = 2, then 2x= 2 -1 = 1, so 2x/-2 = 1/2 . So x= -1/2.

If 2x+1 = -2, then 2x = -2-1 = -3. So 2x/2 = -3/2 = -3/2 = -3/2

**Therefore if 26|2x+1| = 52, then x= 1/2 or x= -3/2.**

For the beginning, we'll discuss 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:

26(2x + 1) = 52

We'll divide by 26:

2x + 1 = 2

We'll subtract 1 both sides, to isolate x to the left side:

2x = 2-1

We'll divide by 2:

**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:

26(-2x - 1) = 52

-2x - 1 = 2

We'll add 1 both sides:

-2x = 3

**x = -3/2**

Since x = -3/2 belongs to the interval of admissible values, (-infinite, -1/2), we'll accept it, too.

**The solutions of the given equation are: {-3/2 ; 1/2}.**