Question 1192806
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The other tutor uses a method that I would not choose; there are two other standard solution methods that I think are easier.<br>
And his final statement that the inequality is a contradiction is not mathematically correct; his statement should be that the solution set to the inequality is the empty set.<br>
First alternative method....<br>
|x-a| can be interpreted as the distance between x and a.  So<br>
|x| = |x-0|
|2-x| = |x-2|<br>
Then the inequality can be read as "the sum of the distances from x to 0 and from x to 2 is less than 2".  Picturing points on a number line, it is easy to see that |x|+|2-x| is EQUAL to 2 for 0 < x < 2; and |x|+|2-x| is greater than 2 everywhere else.  So there is no solution the inequality; its solution is the empty set.<br>
Second alternative method....<br>
The terms |x| and |2-x| mean the behavior of the expression changes at x=0 and x=2.  So analyze the three cases for x less than 0, for x between 0 and 2, and for x greater than 2.<br>
(1) If x is less than 0, then |x| = -x and |2-x| = 2-x.  The inequality is then
(-x)+(2-x)<2
-2x+2<2
-2x<0
x>0<br>
But that "solution" is not in the range of x values we are working with (x less than 0).  So there is no solution for x less than 0.<br>
(2) If x is between 0 and 2, then |x| = x and |2-x| = 2-x; the inequality is then
(x)+(2-x)<2
2<2<br>
That statement is never true; so there is no solution for x between 0 and 2.<br>
(3) If x is greater than 2, then |x| = x and |2-x| = x-2; the inequality is then
(x)+(x-2)<2
2x-2<2
2x<4
x<2<br>
But that "solution" is again not in the range of x values we are working with (x greater than 2).  So there is no solution for x greater than 2.<br>
This shows that there is no solution to the inequality in any of the three intervals; therefore, there is no solution to the inequality -- the solution is the empty set.<br>