Find all numbers x for which |x-1| + |x-2| > 1.
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.
To find the solution to the inequality |x-1| + |x-2| > 1, let's break it down into cases based on the values of x.
Case 1: x ≥ 2
When x is greater than or equal to 2, both x-1 and x-2 are positive or zero. So, the absolute value of x-1 is (x-1) and the absolute value of x-2 is (x-2). Therefore, the inequality becomes:
(x-1) + (x-2) > 1
2x - 3 > 1
2x > 4
x > 2
Since we are already assuming x ≥ 2, there are no additional values of x that satisfy this condition.
Case 2: x < 2
When x is less than 2, x-1 is negative and x-2 is also negative. So, the absolute value of x-1 is -1*(x-1) and the absolute value of x-2 is -1*(x-2). Therefore, the inequality becomes:
(-1)*(x-1) + (-1)*(x-2) > 1
-2x + 3 > 1
-2x > -2
x < 1
Again, since we are already assuming x < 2, there are no additional values of x that satisfy this condition.
Therefore, combining the two cases, we can conclude that there is no value of x that satisfies the inequality |x-1| + |x-2| > 1.