Chapter 1, Problem 10P

To determine

To find: The sketch for the region in the plane that consists of all the points $\left(x,y\right)$ such that $\left|x-y\right|+\left|x\right|-\left|y\right|\le 2$ .

Explanation of Solution

Given:

The given condition is $\left|x-y\right|+\left|x\right|-\left|y\right|\le 2$ .

Calculation:

Consider the given condition is,

  $\left|x-y\right|+\left|x\right|-\left|y\right|\le 2$

Consider the variable $x$ as,

  $\left|x\right|=\{\begin{array}{l}x\text{if}x\ge 0\\ -x\text{if}x<0\end{array}$

Consider the variable $y$ as,

  $\left|y\right|=\{\begin{array}{l}y\text{if}y\ge 0\\ -y\text{if}y<0\end{array}$

Then, consider the difference,

  $\left|x-y\right|=\{\begin{array}{l}x-y\text{if}x-y\ge 0\\ x--y\text{if}x-y<0\end{array}$

Consider the case when $x>0$ and $y<0$ .

  $\begin{array}{l}\left|x-y\right|+\left|x\right|-\left|y\right|\le 2\\ x-y+x-\left(-y\right)\le 2\\ 2x\le 2\\ x\le 1\end{array}$

Consider the case when $y>0$ and $x<0$ .

  $\begin{array}{l}\left|x-y\right|+\left|x\right|-\left|y\right|\le 2\\ -\left(x-y\right)-x-y\le 2\\ -2x\le 2\\ x\ge -1\end{array}$

Consider the case when $y>x$ , $x>0$ and $y>0$ .

  $\begin{array}{l}\left|x-y\right|+\left|x\right|-\left|y\right|\le 2\\ -\left(x-y\right)+x-y\le 2\\ 0\le 2\left(\text{true}\right)\end{array}$

Consider the case when $y<x$ , $x>0$ and $y>0$ .

  $\begin{array}{l}\left|x-y\right|+\left|x\right|-\left|y\right|\le 2\\ \left(x-y\right)+x-y\le 2\\ x-y\le 1\\ \text{y}\ge 1\end{array}$

Consider the case when $y<x$ , $x<0$ and $y<0$ .

  $\begin{array}{l}\left|x-y\right|+\left|x\right|-\left|y\right|\le 2\\ \left(x-y\right)-x-\left(-y\right)\le 2\\ 0\le 2\left(\text{true}\right)\end{array}$

Consider the case when $y>x$ , $x<0$ and $y<0$ .

  $\begin{array}{l}\left|x-y\right|+\left|x\right|-\left|y\right|\le 2\\ -\left(x-y\right)-x-\left(-y\right)\le 2\\ -2x+2y\le 2\\ \text{y}\le 1+x\end{array}$

The graph for all the points in the region that consists of all the points $x,y$ is shown in Figure 1

  

Figure 1