We have to solve |2x + 16| = 24

As 2x + 16 is within the modulus sign, we can have

2x + 16 = 24

=> 2x = 8

=> x = 4

and 2x + 16 = -24

=> 2x = -40

=> x = -20

**There are two solutions for x, x = 4 and x = -20**

To solve the equation |2x+16|=24 remember that

|a| = a for a >= 0

|a| = -a for a < 0

|2x+16|=24

If 2x + 16 >= 0

x >= -8

2x + 16 = 24

2x = 8

x = 4

Note that 4 is greater than -8

If 2x + 16 < 0

x < -8

-(2x + 16) = 24

-2x -16 = 24

-2x = 40

x = -20

Note that -20 < -8

The solution of the equation |2x+16|=24 is 4 and -20

We recall the property of absolute value:

|x| = a>0

We'll have to discuss 2 cases:

1) 2x+16 = 24, if 2x+16>=0 => x belongs to [8;+infinite)

We'll subtract 16 both sides:

2x = 24-16

2x = 8

We'll divide by 2:

x = 4

2) 2x+16 = -24, if 2x+16<0 => x belongs to (-infinite,8)

We'll subtract 16 both sides, to isolate x to the left side:

2x = -16 - 24

2x = -40

We'll divide by 2:

x = -20

**Since both values are in the ****admissible ****intervals, they both become the ****solutions of the ****equation: {-20 ; 4}.**