3 l 5x-3 l + 8 = 11 solve for X

First we will move 8 to the right sides:

==> 3 l 5x-3l = 3

Now divide by 3:

==> l 5x-3 l = 1

Now we have two cases:

case(1):

(5x-3)= 1

==> 5x = 4

**==> x= 4/5**

Case(2):

-(5x-3) = 1

==> -5x + 3 = 1

==> -5x = -2

**==> x= 2/5**

**==> x= { 4/5, 2/5}**

We'll solve the module in ths way:

3 l 5x-3 l + 8 = 11

We'll subtract 8 both sides first :

3 l 5x-3 l = 11-8

3 l 5x-3 l = 3

We'll divide by 3:

l 5x-3 l = 1

We'll get 2 cases to solve:

1) We'll impose the constraint of absolute value:

5x - 3 for 5x - 3>=0

5x>=3

x>=3/5

Now, we'll solve the equation:

5x - 3 = 1

We'll add 3 both sides:

5x = 4

x = 4/5

The value of x belongs to the interval of admissible values:

[3/5 , +inf.)

2) -5x + 3 for 5x - 3<0

5x<3

x<3/5

Now, we'll solve the equation:

-5x + 3 = 1

We'll subtract 3 both sides:

-5x = -2

We'll divide by -5:

x = 2/5

Since the value of x belongs to the interval of admissible values, x = 2/5 is also a root of the given equation.

**The roots of the equation are: {2/5 ; 4/5}.**

3|5x-3| +8 = 11.

Subtract 8 from both sides:

3|5x-3| = 11-8 = 3.

3|5x-3| = 3

Divide 3 :

|5x-3| = 3/3 = 1.

Therefore , when 5x-3 > 0 , |5x-3| = 1 implies 5x-3 = 1 , Or 5x = 4. Or x = 4/5.

When 5x-3 < 0, then |5x-3| = 1 implies 3-5x = 1. Or 3-1 = 5x . Or 5x = 2. Or x = 2/5.

Therefore the solutions are x = 4/5 . Or x= 2/5.