# How to solve $\lvert{x}\rvert - \lvert{2+x}\rvert = x$?

How do I solve the following equation? $$\lvert{x}\rvert- \lvert{2+x}\rvert= x$$

I was thinking about dividing it into 4 cases: plus plus, plus minus, minus plus and minus minus. What is the best way to solve this?

$$|x|-|x+2|=\left\{\begin{matrix} -x+(x+2)& \;\;, x\leq -2 \\\\ -x-(x+2)&\;\;, -2<x<0 \\\\ x-(x+2) &\;\; ,x\geq 0 & \end{matrix}\right.$$

$\bullet\;$ If $x\leq -2\;,$ Then equation convert into $-x+x+2=x\Rightarrow x=2\;\;(\bf{False})$

$\bullet\;$ If $-2<x< 0\;,$ Then equation convert into $\displaystyle -x-x-2=x\Rightarrow x=-\frac{2}{3}\;\;(\bf{True})$

$\bullet\;$ If $x\geq 0\;,$ Then equation convert into $x-x-2=x\Rightarrow x=-2\;\;(\bf{False})$

So our solutions are $$\displaystyle x= -\frac{2}{3}$$

• But if I place $x=2$ I get $-2=2$ which is false Oct 28, 2015 at 16:59
• Your made a sign error for the case $x \ge 0$.
– Ojas
Oct 28, 2015 at 16:59
• Thanks Ojas I have edited it. Oct 28, 2015 at 17:02
• Donna I have edited my answer. Oct 28, 2015 at 17:02

Another way is to use $|x| = \sqrt{x^2}$. If $|x| - |2+x| = x$ then \begin{align*} \sqrt{x^2} - \sqrt{(x+2)^2} &= x\\ x^2 + (x+2)^2 -2\sqrt{x^2}\sqrt{(x+2)^2} &= x^2\\ (x+2)^2 &= 2\sqrt{x^2} \sqrt{(x+2)^2}\\ (x+2)^4 &= 4x^2(x+2)^2 \end{align*} So either $x= -2$, which doesn't satisfy the original equation, or else $$(x+2)^2 = 4x^2$$ which is a quadratic equation in $x$. This has the roots $x=2$ and $x=-2/3$. Now you just have to check whether either of these satisfies the original equation, and only $x=-2/3$ does, so this is the only solution.

HINT: Consider the 3 cases:

1. $x<-2$
2. $-2\le x<0$
3. $x\ge 0$
• Ok but why? How did you get there? Oct 28, 2015 at 16:43
• Use the definition: $|x|=x$ when $x\ge 0$ and $|x|=-x$ when $x< 0$ Oct 28, 2015 at 16:44
• Can you tell me what are the steps to solve this equation? I want an algorithm Oct 28, 2015 at 16:50
• The hint above is the first step in the algorithm. Have a think about why those three cases are important in your problem. Then for each case replace the absolute value with a plus or minus value as appropriate, then solve and see if the answer is in the domain for that case. Oct 28, 2015 at 16:53
• Whenever you have an expression inside a mod sign, you will realise that there are 2 signs of the expression inside the modulus for different values of the variable and hence different solutions of the variable. What you need to do is that lift the modulus signs and then work out the equation. But then all the modulus signs have two values as per definition. So you have to consider such cases and then solve the problem. Oct 28, 2015 at 16:55

$$\lvert{x}\rvert + 2 \ge\lvert{2+x}\rvert$$ $$\therefore x = \lvert{x}\rvert- \lvert{2+x}\rvert \ge -2$$ $$\therefore x+2 \ge 0$$ Thus, $$\lvert{x}\rvert- \lvert{2+x}\rvert= x$$ becomes$$\lvert{x}\rvert = 2x + 2$$ Now, just consider two cases. $x \lt 0$ and $x \ge 0$.

If $x<-2$ then $x + 2 < 0$. So $\lvert x \rvert - \lvert 2 + x \rvert = - x + (2+x)$. So you really just want to solve $$-x + (2+x) = x.$$ Here the solution is $x = 2$. This contradicts $x < -2$. So there are not solutions for $x<-2$.

Now if $-2\leq x < 0$, then $\lvert x \rvert = -x$ and $\lvert 2+x\rvert = 2 + x$.

If $x \leq 0$, then $\lvert x\rvert = x$ and $\lvert 2 +x\rvert = 2+x$.

All of this comes down to the fact that $$\lvert \text{something}\rvert = \begin{cases}\text{something} & \text{if } \text{something} \geq 0 \\ -\text{something} & \text{if } \text{something} < 0 \end{cases}$$