Solve absolute value inequality with x and y I have difficulties to solve this system of inequalities. Normally you would solve it like a normal system of equations just writing $\le$ or $\ge$ instead of $=$. But these contain an absolute value which confuses me. The system I want to solve is this:
$$
\left\{ 
\begin{array}{c}
|x-y| \le 1 \\ 
|x+y| \le 1
\end{array}
\right. 
$$
I would have written it like that
$$
\left\{ 
\begin{array}{c}
-1 \le x-y \le 1 \\ 
-1 \le x+y \le 1
\end{array}
\right. 
$$
And by adding the inequalities I would obtain $-1 \le x \le 1$. I can’t get $y$ though.
I don’t understand how my textbook got a third inequality out of the two initial ones
$$
\left\{ 
\begin{array}{c}
-1 \le x-y \le 1 \\
-1 \le y-x \le 1 \\
-1 \le x+y \le 1
\end{array}
\right. 
$$
But with that they can solve the system like so: $(1)+(3): -1 \le x \le 1$ and $(2)+(3): -1 \le y \le 1$.
 A: Let $S$ be the set of points in the cartesian plane which solve the system. In the last line of your question you found that $S$ is contained (and not equal) to the square $[-1,1]\times [-1,1]$.
Note that $|x-y| \le 1$ is the strip between the lines $y=x-1$ and $y=x+1$.
$|x+y| \le 1$ is the strip between the lines $y=-x-1$ and $y=-x+1$. What is the intersection of these two strips? Make a drawing and you will find $S$.
A: The third equation comes from multiplying by a negative. The inequality $-1 \leq x - y \leq 1$ is equivalent to multiplying it through by a constant, like -1, yielding the inequality $1 \geq y -x \geq -1$ (you must flip the signs when multiplying by a negative). If we want to keep the original phrasing and use a less than or equal to sign, we just shift the order, and the inequality becomes $ -1 \leq y-x \leq 1$
It's really similar to solving a system of equations where you would multiply the equation by a constant so that adding the equations cancels out a term. We did the exact same thing here, and the constant was $-1$.
