# 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 $(2)+(3): -1 \le y \le 1$.

• 'inequality', not 'equation' (dotted a few places throughout the question) – Shuri2060 Jul 1 '17 at 16:47

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 $-1$.
• It's just like how the equations $2x = 4$ and $x = 2$ say the same thing, the only difference is that the second equation is the first equation multiplied through by one half. The second equation is more useful to us, because generally we want to know what x is, not what twice of x is. When solving these systems using this technique, whether they're inequalities or not, we're rewriting the the system so it can cancel, so that we can ultimately determine what each variable is on its own. – Cordello Jul 1 '17 at 17:01
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 $S$.