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

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

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  • $\begingroup$ But why do they do that? Multiplying by a constant doesn’t give me new information for the system...So it shouldn’t help me solve it?! And yet it helped them. $\endgroup$ – philmcole Jul 1 '17 at 16:47
  • $\begingroup$ You're right, it preserves all the information of the system. The original inequality and the inequality multiplied by a constant say the same thing, you've just changed it into a form where it cancels nicely. To solve for x, you had the luxury of them canceling nicely already, which will not always be the case. So for y, you had to rewrite the system in an equivalent form to get it to cancel. They have the same information, you just manipulate the information into a more useful form. $\endgroup$ – Cordello Jul 1 '17 at 16:57
  • $\begingroup$ 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. $\endgroup$ – Cordello Jul 1 '17 at 17:01
  • $\begingroup$ Ok thanks, so I can get y out of the same set of equations which I already used to get x? $\endgroup$ – philmcole Jul 1 '17 at 17:38
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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 $S$.

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