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Solve the following system of equations:

$|x|+|y|=2$ and $y=x+1$, where $x$ is a real number.

Approach:

I substituted $y$ in equation $1$, so:

$$ \ \ \ \ \ \ \ \ \ \ |x|+|x+1|=2$$

$$1 \ \ \ \ \ \ \ \ \ \ 1$$

$$2 \ \ \ \ \ \ \ \ \ \ 0$$

$$0 \ \ \ \ \ \ \ \ \ \ 2$$

These are the total possibilities I think, because $|x|, |x+1| \ge 0$.

Then I made cases,

Case $1$: $|x|=1$, so, $x = \pm 1$

and $|x+1| = 1$, so $x+1=\pm 1$ so, $x = 0,-2$.

But no value of $x$ is matching, so this case gets rejected.

Case $2$: $|x|=2$, so, $x = \pm 2$

and $|x+1| = 0$, so $x=-1$.

But no value of $x$ is matching, so this case also gets rejected.

Case $3$: $|x|=0$, so, $x = 0$

and $|x+1| = 2$, so $x+1 = \pm 2$ and so, $x=1,-3$.

Here also, no value of $x$ is matching. So no solution exists.

Is this solution correct? Please confirm. If there is a shorter method to approach the question, please share it.

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    $\begingroup$ Is $x\in\Bbb R$? In that case $x=\frac{1}{2}$ would also work $\endgroup$
    – Patricio
    May 2 at 8:12
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    $\begingroup$ It is a lot easier to find the solutions graphically. Are you allowed to do that? $\endgroup$
    – DatBoi
    May 2 at 8:13
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    $\begingroup$ @DatBoi, yes graphical soln is allowed. $\endgroup$
    – user907745
    May 2 at 8:14
  • $\begingroup$ @DatBoi Yes, as intersections of a square (first equation) and a straight line (second equation). $\endgroup$
    – Jean Marie
    May 2 at 8:14
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    $\begingroup$ @lonestudent I did, thanks. $\endgroup$
    – user907745
    May 2 at 9:13
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enter image description here

Image courtesy $-$ Martund.

$|x|+|y|=2$ is square on graph paper (you will get it by making cases).

and $y=x+1$ is line.

This is the shortest (not only) method I know.

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    $\begingroup$ I suggest you to add this picture in your answer (as a smoother version of yours). I've made it using Microsoft Paint. $\endgroup$
    – Martund
    May 3 at 8:11
  • $\begingroup$ @Martund always feel free to edit my answers and questions and thanks for help. $\endgroup$
    – Jay
    May 3 at 8:49
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We know that $$|x-y|\le|x|+|y|$$ where equality holds iff $x$ and $y$ are of opposite sign. In this case we know that equality doesn't hold, so $x$ and $y$ are of same sign, so the first equation becomes $|x+y|=2$ and second is $y-x=1$. Making cases for positive and negative sign in first equation, we get the required solutions: $$\left(\frac12,\frac32\right)\quad\text{and}\quad\left(-\frac32,-\frac12\right)$$

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  • $\begingroup$ thanks for your answer (I ran out of upvotes I'm sorry about that). Could you explain how we know $x$ and $y$ are of same sign... $\endgroup$
    – user907745
    May 2 at 9:13
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    $\begingroup$ @Crease, if they were of opposite sign equality would have held in the first equation. But we know that the LHS is $1$ and RHS is $2$ for that equation, so $x$ and $y$ are of the same sign. $\endgroup$
    – Martund
    May 2 at 9:16
  • $\begingroup$ Oh yes, thanks :) $\endgroup$
    – user907745
    May 2 at 9:20
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HINT

  • If $x≥0$, then $y=x+1>0$

$$\begin{cases} x+y=2 \\ y=x+1 \end{cases}$$

Then, we need

  • If $-1≤x<0, ~y≥0 $

$$\begin{cases} y-x=2 \\ y=x+1 \end{cases}$$

  • If $x<-1, ~y<0$

$$\begin{cases} -x-y=2 \\ y=x+1 \end{cases}$$

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  • $\begingroup$ Of course, for the second case, the solution doesn't exist. $\endgroup$ May 2 at 11:35
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One nice way to think about it is to see |x-a| as the distance between x and a. Visually, it's pretty clear only x=-1.5 and x=0.5 are solutions.

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You only consider integers, hence you might left out non-integer solutions.

We have $$|x|+|x+1|=2$$

$$|x-0|+|x-(-1)|=2$$

So the solution for $x$ is when the sum of distance from $0$ and the distance from $-1$ is exactly $2$.

Clearly, the point can't be inside $[-1, 0]$.

If $x$ is positive, then we have $2x+1=2$, hence $x=\frac12$.

Similarly, if $x<-1$, by symmetry, $x=-1-\frac12=-\frac32$.

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