- #1

brotherbobby

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- Homework Statement
- Solve ##|x|+|x-2|=2##

- Relevant Equations
- 1. For a given function ##f(x)##, if ##|f(x)|<a \Rightarrow -a < f(x) < a##

2. For a given function ##f(x)##, if ##|f(x)|>a\Rightarrow f(x)<-a\; \text{or}\; f(x)>a##

(I could solve the problem but could not make sense of the solution given in the text. Let me put my own solutions below first).

(1) ##\mathbf{x \ge 2}## : In this case we have given ##|x|+|x-2|=2\Rightarrow x+x-2 = 2\Rightarrow 2x=4\Rightarrow \boxed{x=2}## (as the answer ##x=2## lies within the selected range in which the equation is solved).

(2) ##\mathbf{0\le x<2}## : For this case, the equation: ##|x|+|x-2|=2\Rightarrow x+2-x = 2 \Rightarrow 2=2## which is always true. Hence the selected range is a solution : ##\boxed{0\le x<2}##.

(3) ##\mathbf{x<0}## : In this case, the given equation : ##|x|+|x-2|=2\Rightarrow -x+2-x = 2\Rightarrow 2x=0\Rightarrow x=0##. This solution has to be rejected for it doesn't lie in the selected range.

Hence, the solution to the problem : ##\boxed{0\le x\le 2}##. ##\Large{\checkmark}##, as it agrees with the answer in the text.

It is the solution that I am struggling to follow. Namely on the two steps marked as

In

**I copy and paste the problem to the right as it appears in the text.
1. Problem Statement : ****2. My attempt :**There are three "regions" where ##x## can lie.(1) ##\mathbf{x \ge 2}## : In this case we have given ##|x|+|x-2|=2\Rightarrow x+x-2 = 2\Rightarrow 2x=4\Rightarrow \boxed{x=2}## (as the answer ##x=2## lies within the selected range in which the equation is solved).

(2) ##\mathbf{0\le x<2}## : For this case, the equation: ##|x|+|x-2|=2\Rightarrow x+2-x = 2 \Rightarrow 2=2## which is always true. Hence the selected range is a solution : ##\boxed{0\le x<2}##.

(3) ##\mathbf{x<0}## : In this case, the given equation : ##|x|+|x-2|=2\Rightarrow -x+2-x = 2\Rightarrow 2x=0\Rightarrow x=0##. This solution has to be rejected for it doesn't lie in the selected range.

Hence, the solution to the problem : ##\boxed{0\le x\le 2}##. ##\Large{\checkmark}##, as it agrees with the answer in the text.

**I copy and paste the solution in the text given to the right.
3. The text's attempt : **It is the solution that I am struggling to follow. Namely on the two steps marked as

**1**and**2**.In

**1**, the author is tacitly assuming that ##x>0## and that ##x<2##. What is the reason for this assumption? There are three broad regions where ##x## can lie, as I have shown in my solution above. Why take the one on the middle, as he has done?**2**is more confusing. How can the expression ##x-(x-2)=2## become ##x(x-2)\le 0##?**A hint or suggestion referring to the author's solution would be welcome.**I admit that he obtains the same answer as I do.