Find all numbers x for which |x-1| + |x-2| > 1.

Question

I know the answer is that there exists no such x, but I need someone to explain to me the steps as to figuring out why that is the case.

Answer 1

Between 1 and 2, you have

y = (x-1) + -(x-2) = 1

everywhere else, y > 1

for x < 1, y = -(x-1) + -(x-2) = 3-2x
for x > 2, y = (x-1) + (x-2) = 2x-3

Answer 2

| x - 1 | =

x - 1

OR

- | x - 1 | = 1 - x

| x - 2 | =

x - 2

OR

- | x - 2 | = 2 - x

4 combinations are possible.

1. combination:

x - 1 + x - 2 > 1

2 x - 3 > 1

2 x > 1 + 3

2 x > 4 Divide both sides by 2

x > 2

2. combination:

x - 1 + 2 - x > 1

1 > 1

That is not true and that is not solution.

3. combination:

1 - x + x - 2 > 1

- 1 > 1

Again, that is not true and that is not solution.

4. combination:

1 - x + 2 - x > 1

- 2 x + 3 > 1

- 2 x > 1 - 3

- 2 x > - 2 Divide both sides by - 2

When you multiply or divide both sides of inequality by a negative number, inequality change the direction. So:

x < 1

Solutions are :

x < 1

and

x > 2