# solve the module:3 I 5x-3 I + 8 = 11

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Isolating the absolute value to one side, yields:

`3|5x - 3| = 11 - 8 => 3|5x - 3| = 3 => |5x - 3| = 1`

You need to use the definition of absolute value, such that:

`|5x - 3| = {(5x - 3, 5x - 3 >= 0),(3 - 5x, 5x - 3 < 0):}`

`|5x - 3| = {(5x - 3, x >= 3/5),(3 - 5x, x < 3/5):}`

`5x - 3 = 1 => 5x = 3 + 1 => 5x = 4 => x = 4/5 > 3/5` valid

`x = 0.8`

`3 - 5x = 1 => -5x = -3 + 1 => -5x = -2 => x = 2/5 < 3/5` valid

`x = 0.4`

**Hence, evaluating the absolute value equation yields **`x = 0.4, x = 0.8.`

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}.