How many real solutions (x,y) from |x-y| + |x+y| = 1 ?

I really wonder how to find it.

My attempt:

I think I need to separate this problem into some cases: First case: for |x-y| >0

we got:

x-y + |x+y| =1

|x+y|= 1-x+y

for |x+y|>0, we got:

x+y= 1-x+y

x= 1/2

Second case, for |x-y| <= 0

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

|x+y|= 1+x-y

For |x+y| >0, we got:

x+y= 1+x-y

2y= 1

y= 1/2

But, for every cases, I think I can only get one result for each variable, such that x or y.

How to find the solutions set for (x,y) by using any cases?


  • $\begingroup$ $|x-y|<0$ is not possible $\endgroup$ – graydad Oct 4 '14 at 17:15
  • $\begingroup$ Is "how many" really the question you want to ask? Are you looking for the cardinality of the set of solutions? $\endgroup$ – André Nicolas Oct 4 '14 at 17:31
  • $\begingroup$ See math notation guide. $\endgroup$ – user147263 Oct 4 '14 at 19:05

If we square both sides of the equation

$$|x - y| + |x + y| = 1$$

we obtain

$$(x - y)^2 + 2|x^2 - y^2| + (x + y)^2 = 1$$

which simplifies to

$$2x^2 + 2y^2 + 2|x^2 - y^2| = 1$$

Case 1: If $|x| \geq |y|$, then we obtain

$$4x^2 = 1$$

so $x = \pm \dfrac{1}{2} \Rightarrow |y| \leq \dfrac{1}{2}$.

Case 2: If $|x| < |y|$, then we obtain

$$4y^2 = 1$$

so $y = \pm \dfrac{1}{2} \Rightarrow |x| < \dfrac{1}{2}$.

Note that the solutions $\left(\pm \dfrac{1}{2}, \pm \dfrac{1}{2}\right)$ were accounted for in Case 1.

Hence, the solution set is

$$S = \left\{(x, y) \in \mathbb{R}^2 \bigg|~x = \pm \frac{1}{2}, -\frac{1}{2} \leq y \leq \frac{1}{2}\right\} \bigcup \left\{(x, y) \in \mathbb{R}^2 \bigg|~y = \pm \frac{1}{2}, -\frac{1}{2} < x < \frac{1}{2}\right\}$$


If $(x, y)$ is a solution, then so are $(-x, -y)$ and $(y, x)$. Thus WLOG we can assume that $x \geq |y|$. Clearly then, $x-y \geq 0$ and $x+y \geq 0$ and so we can remove all $|\cdot |$s.

$$(x - y) + (x + y) = 1 \iff x = \frac{1}{2} \geq |y|$$

Thus our set of solutions is $S = \big\{(\pm\frac{1}{2}, r),\;(r, \pm\frac{1}{2})\:|\:r\in [-\!\frac{1}{2}, \frac{1}{2}]\big\}$, and then the "how many" is $|S| = 4\big|[-\!\frac{1}{2}, \frac{1}{2}]\big| = \aleph_1$.

  • 1
    $\begingroup$ Well, $|S|=\mathfrak{c},$ anyway. $\endgroup$ – Cameron Buie Oct 4 '14 at 17:39
  • 1
    $\begingroup$ It might be good to use $\geq$ instead of $>$ here, since $\left(\frac{1}2,\frac{1}2\right)$ is a solution and if $z\geq 0$ it still holds that $|z|=z$. $\endgroup$ – Milo Brandt Oct 4 '14 at 17:40
  • $\begingroup$ Correct. Thanks :) $\endgroup$ – Ben Frankel Oct 4 '14 at 17:41

First of all you can observe that exchanging $x$ and $y$ produces the same equation, so if $(a,b)$ is a solution, also $(b,a)$ is.

Thus it's not restrictive to assume $x\ge y$, just remember to add back the solutions with terms reversed. Thus $|x-y|=x-y$ and we have to consider the two cases

  1. $x+y\ge0$

  2. $x+y<0$.

Case 1 The equation is $x-y+x+y=1$ or $2x=1$ that means $x=1/2$. There's no restriction on $y$ apart from $y\ge x$ and $y\ge -x$, that is $-1/2\le y\le 1/2$.

Case 2 The equation is $x-y-x-y=1$ or $2y=-1$, which gives $y=-1/2$. There's no restriction on $x$ apart from $x\ge y$ and $x<-y$ or $-1/2\le x<1/2$.

The solutions of the first case are the pairs $$ (1/2,t)\quad \text{for $-1/2\le t\le 1/2$} $$ so also the pairs $$ (t,1/2)\quad \text{for $-1/2\le t\le 1/2$} $$ are solutions.

The solutions of the second case are the pairs $$ (t,-1/2)\quad \text{for $-1/2\le t< 1/2$} $$ so also the pairs $$ (-1/2,t)\quad \text{for $-1/2\le t< 1/2$} $$ are solutions.


I believe you mean your cases to be $x-y>0$ and $x-y\le 0.$ Still, as you've seen, that's not getting you the results you want.

One alternative is to rearrange the equation to isolate one of the absolute value expressions, then square both sides. This allows you to use the fact that $|u|^2=u^2$ for all real numbers $u.$ You will still have an absolute value expression, so rearrange and square again. Now you have no more absolute value expressions, and can try to determine possible solutions from there. Unfortunately, this is still fairly intractable to solve.

A better alternative is to consider the cases $x+y\ge0$ and $x+y<0.$ In the first case, the equation becomes $$x+y+|x-y|=1.$$ Here's where the slick trick comes in: multiplying both sides of the equation by $\frac12$ gives us $$\max(x,y)=\frac12.$$ Can you see why?

In the second case, the equation becomes $$-x-y+|x-y|=1,$$ or equivalently, $$x+y-|x-y|=-1.$$ Again, we multiply by $\frac12,$ giving us $$\min(x,y)=-\frac12.$$ Do you see why?

Putting the cases together, the solution set is a square, which can be described by $\max\bigl(|x|,|y|\bigr)=\frac12.$

Let me know if you have any questions.


One note on your proof is that you should say the case is where $x+y\leq 0$ not where $|x+y|<0$. The absolute value is always non-negative - you're interested in whether, in $|z|$ the value $z$ is positive. And if $z$ is zero then $|z|=z=-z$, so you might as well use $\leq$ instead of $<$.

What you should interpret getting a result for only one variable is that, for instance, where you show that if $x-y\geq 0$ and $x+y\geq 0$, then $x=\frac{1}2$, then the equation $|x-y|+|x+y|=1$ holds for any pair $(\frac{1}2,y)$ where $\frac{1}2-y\geq 0$ and $\frac{1}2+y\geq 0$, or equivalently, when $\frac{-1}{2}\leq y \leq \frac{1}2$. Essentially, it means that the other value can vary wherever it pleases, so long as the conditions necessary for that case still hold. You can do this to all your cases.


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