8 + l 1-x/2 l > or = 10?

Answer 1

8 + |1 - x/2| >= 10
|1 - x/2| >= 2

1 - x/2 >=2
OR
1 - x/2 <= -2

1 - x/2 >=2
-x/2 >= 1
x/2 <= -1
x <= -2


1 - x/2 <= -2
-x/2 <= -3
x/2 >= 3
x >= 6

x = (-inf, -2) U (6, inf)

Answer 2

you need to solve for the x
but with the abs value you end up with two answers

8 + (1-x/2) >=10
and
8 - (1-x/2) >=10

1) 8 + (1-x/2) >=10
(1-x/2) > = 2
-x/2 >= 1
-x >=2
x <= -2

2) 8 - (1-x/2) >=10
-(1-x/2) > = 2
-1 +x/2 >= 2
x/2 > = 3
x >=6

Verify by substitution

if x = -2
8 + |1-(-2)/2| = 8 + |1+1| = 10 (checks out)
if x = -4
8 + |1-(-4)/2| = 8 + |1+2| = 11 (checks out)

if x = 6
8 + |1-(6)/2| = 8 + |1-3| = 8 +2 = 10 (checks out)
if x = 8
8 + |1-(8)/2| = 8 + |1-4| = 11 (checks out)

let x = 0
8 + |1-(0)/2| = 8 + |1| = 9 (Not >=10)

So your answer is x<=-2 and x >=6

Answer 3

8 + | 1 - x/2 | ≥ 10
|1 - x/2| ≥ 2

Case 1
1 - x/2 ≥ 2
- 1 ≥ x/2
x ≤ - 2

Case 2
1 - x/2 ≤ - 2
- x / 2 ≤ - 3
x/2 ≥ 3
x ≥ 6

Answer is x ≤ - 2 , x ≥ 6