1. solve: |x| = 5
{-5}
{-5, 5}
{5}
{-1, 5}
2. solve: | m + 8| = 16
{-24, 8}
{8, 16}
{8}
{-24}
3. solve: |2n - 3| = 5
{1, -4}
{-1, 1}
{-1, 4}
{1, 4}
4. solve: |x/3 + 1| = 6
{-21, 15}
{-7, 15}
{-15, 21}
{7, 15}
5. solve |2 - 3x| = 7.
{-3, 5/3}
{-3, 15}
{-5/3, 3}
{-15, 3}
6. solve |2y - 2| = y.
{2/3, 2}
{1/3, 2}
{2, 5}
{2, 3}

Answers: 

1)  [B]:  {-5, 5} 
2)  [A]:  {-24, 8} 
3)  [C]:  {-1, 4} 
4)  [A]:  {-21, 15}
5)  [C]:  {-5/3, 3}
6)  [A]:  {⅔, 2}

Explanation:

2)  "Solve: | m + 8| = 16 ";  Assuming we are to solve for "m"; we would use:

"Case 1" and "Case 2" scenarios;

 since we need to find the value for "m" when:

Case 1:   (m + 8) = 16; AND:

Case 2:  - (m + 8) = 16 .

→ Let's solve, starting with Case 1:

m + 8 = 16;  → Subtract "8" from EACH SIDE of the equation;

→ m + 8 − 8 = 16 − 8 ; →  To get: →  m = 8

→ Let us solve for "Case 2" :

- (m + 8) = 16 ;

Note the distributive property of multiplication:

→ a*(b + c) = ab + ac ;

→ As such: - (m + 8) = 16 ;

Consider this equation as:

- 1(m + 8) = 16 ;

→ Note the  "-1" is implied, since anything multiplied by "1" is that same thing. The "negative" in the "negative 1" comes from the "negative sign" already present.

→  - 1(m + 8) = 16 ;   →  (-1*m) + (-1*8) = 16 ; 

→  -1m + (-8) = 16 ; →  Rewrite as: -1m − 8 = 16 ; 

→ Add "8" to EACH SIDE of the equation ; →  -1m − 8 + 8 = 16 + 8 ;
→  to get: -1m = 24 ;

→ Divide EACH SIDE of the equation by "-1"; to isolate "m" on one side of the equation; & to solve for "m" ;  →  -1m / -1 = 24 / -1 ; → m = -24

So we have: m = 8; and m = -24 ; which is: "Answer choice: [A]: {-24, 8}" .

3) Given: |2n - 3| = 5 ; Solve for "n" ; 

Case 1:  2n − 3 = 5 ; 

→ Add "3" to EACH side of the equation;  2n − 3 + 3 = 5 + 3 ; → 2n = 8 ; 
    →  Now, divide EACH SIDE of the equation by "2"; to isolate "n" on one side of the equation; and to solve for "n";     → 2n / 2 = 8 / 2 ;  → n = 4 .

Case 2: - (2n − 3) = 5 ;

Simplify: - (2n − 3);  → - 1(2n − 3)  = (-1*2n) − (-1*3) = -2n − (-3) = -2n + 3 ;

→ Rewrite the equation, "- (2n − 3) = 5 " ;  as:
→  -2n + 3 = 5;   → Now, subtract "3" from EACH side of the equation:
→  -2n + 3 − 3 = 5 − 3 ; → -2n = 2 ;

→ Divide EACH SIDE of the equation by "-2"; to isolate "n" on one side of the equation; and to solve for "n" ;  → -2n / -2 = 2 / -2 ; → n = -1 . 
→ So, n = 4, AND -1; which is: Answer choice: [C]:  {-1, 4} .

4)  Given: |x/3 + 1| = 6 ; Solve for "x";

Case 1: (x/3) + 1 = 6; Subtract "1" from EACH side of the equation;
 
→ (x/3) + 1 − 1 = 6 − 1 ; → (x/3) = 5 ; → x = 5*3 = 15
 
Case 2:  - [ (x/3) + 1] = 6;  → -1 [(x/3) + 1] = 6 ; → [-1*(x/3) ] + (-1*1) = 6 ;  

→   - (x/3) + (-1) = 6 ;  →  - (x/3) − 1 = 6 ; 
→ Add "1" to EACH side of the equation:  
→ - (x/3) − 1 + 1 = 6 + 1 ;  →  - (x/3) − 1 + 1 = 6 + 1 ; 
   →  - (x/3) = 7 ↔  -x / 3 = 7;  -1x = 7*3 ; →  -1x = 21;

→ Divide EACH SIDE of the equation by "-1"; to isolate "x" on one side of the equation; & to solve for "x" ;

→ -1x / -1 = 21 / -1 ; → x = -21;  
So, x = 15, AND -21; which is:  Answer choice: [A]:  {-21, 15} .

5)  Given: |2 − 3x| = 7 ; Solve for "x"; 

Case 1:  (2 − 3x) = 7

Given: 2 − 3x = 7 ;  → Subtract "2" from EACH side of the equation;

→ 2 − 3x − 2 = 7 − 2 ; → to get: -3x = 5 ; 

→ Now, divide EACH side of the equation by "-3"; to isolate "x" on one side of the equation; and to solve for "x" ;   →  -3x / =3 = 5/ -3  = - (5/3).

→Case 2:  -(2 − 3x) = 7 ; Simplify,

Note: The distributive property of multiplication:
 → a*(b + c) = ab + ac ; AND:  → a*(b − c) = ab − ac ;

→  - (2 − 3x) = 7 ; →  -1(2 − 3x) = 7 ;  → (-1*2) − (-1*3x) = 7 ;

→  - 2 − (-3x) = 7 ; → - 2 + 3x = 7 ; → Add "2" to EACH side of the equation ;
→  - 2 + 3x + 2 = 7 + 2 ;  → 3x = 9 ; 

→ Now, divide EACH side of the equation by "3"; to isolate "x" on one side of the equation; and to solve for "x" ;   →  3x / 3 = 9 / 3 ;  x = 3; 

So, x = - 5/3; AND 3;  which is: Answer choice: [C]: {-5/3, 3} .

6)  Given: |2y − 2| = y ; Solve for "y";

Case 1:  2y − 2 = y; 

→ Subtract "y" from EACH SIDE of the equation;

→ 2y − 2 = y ;  → 2y − 2 − y = y − y ; → y − 2 = 0 ;

→ Add "2" to EACH SIDE of the equation; to isolate "y" on one side of the equation; & to solve for "y" ;  → y − 2 + 2 = 0 + 2 ; → y = 2

Case 2:  - (2y − 2) = y

→ Simplify: - (2y − 2);  →  -1(2y − 2) = (-1*2y) − (-1*2) = -2y - (-2) = -2y + 2 ;

→ Rewrite:  - (2y − 2) = y ; as: → -2y + 2  = y ; 

 → Add "2y" to EACH SIDE of the equation:
           →  -2y + 2 + 2y = y + 2y ; → 2 = 3y ;

→ Now, divide EACH side of the equation by "3"; to isolate "y" on one side of the equation & to solve for "y" ;
→ 2 / 3 = 3y / 3 ;  → ⅔ = y ↔ y = ⅔ ; so y = 2, AND ⅔; which is: 
→ Answer choice: [A]:  {⅔, 2}

If gravity is one third on mars then weight also one third on mars