The 2 absolute equation equality? find the max and min of x?

[Helly123]

Homework Statement

Homework Equations

The Attempt at a Solution

2x - 1 >= 0
x >= 1/2

x - 2 >= 0
x >= 2

for x <1/2
then 2x - 1 and x - 2 are negative
solve : -(2x-1) - (x-2) = 2
-3x + 3 = 2
x = 1/3

for 1/2 <= x < 2
the (2x - 1) positive, (x-2) negative
solve : 2x -1 -x + 2 = 2
x = 1

for x>= 2
both 2x -1 and x -2 = positive
2x - 1 + x - 2 = 2
3x = 5
x = 5/3

is my answer right? then the max of x is 5/3? and the min is 1/3?

 

[Mark44]

Homework Statement

View attachment 205815

Homework Equations

The Attempt at a Solution

2x - 1 >= 0
x >= 1/2

x - 2 >= 0
x >= 2

for x <1/2
then 2x - 1 and x - 2 are negative
solve : -(2x-1) - (x-2) = 2
-3x + 3 = 2
x = 1/3

for 1/2 <= x < 2
the (2x - 1) positive, (x-2) negative
solve : 2x -1 -x + 2 = 2
x = 1

for x>= 2
both 2x -1 and x -2 = positive
2x - 1 + x - 2 = 2
3x = 5
x = 5/3

is my answer right? then the max of x is 5/3? and the min is 1/3?

These are the values I get, as well.

 

[Helly123]

These are the values I get, as well.

so what's the max and min of x? because the key answer not 1/3 and 5/3. but 1/3 and 1

 

[mjc123]

Your third case is self-contradictory. On the assumption that x >= 2, you get x = 5/3, which is < 2. So this case is ruled out, and the answers are 1/3 and 1.

 

[Mark44]

so what's the max and min of x? because the key answer not 1/3 and 5/3. but 1/3 and 1

Your third case is self-contradictory. On the assumption that x >= 2, you get x = 5/3, which is < 2. So this case is ruled out, and the answers are 1/3 and 1.

Yes, @mjc123 is correct. I wasn't careful enough in checking my solutions. The maximum value is x = 1.