Is this true :$-3\leq x \leq5, -2\leq y\leq -1 \implies -1\leq x-y\leq 6$?

Question

let $x , y$ be a real numbers with $-3\leq x \leq5$ and $-2\leq y\leq -1$ , I ask if the range of $x-y$ is $[-1,6]$ or $[-1,1]$ using two cases negative $x$ and positive $x$ and takin g sup is $1$ ?

Answer 1

You should end up with:

$$-2\le x-y \le 7$$

When you negate the $y$ inequality, you have:

$$2\le -y \le 1$$

which is obviously wrong, you should have:

$$2 \ge -y \ge 1$$

and reversing gives:

$$1 \le -y \le 2$$

Add this to the $x$ inequality and you get the correct answer.

Answer 2

No, you cannot subtract inequalities, only add them, and changing signs in an inequality reverses it.

Here a way to go: $$−2≤y≤−1\iff 1\le -y\le 2$$ so, adding the latter to the inequalities with $x$ yields $$-3+1\le x+(-y)=x-y\le 5+2,\quad\text{ i.e. }\quad -2\le x-y\le 7.$$

Answer 3

Two in-equations in the same direction can be added but they can never be subtracted. Let us take two parts of four given in-equations: $x \le 5~ and ~ -2 \le y~$ we can add the two to get $x-2 \le 5+y \implies x-y \le 7.$ Next, let us take other two parts as $-3 \le x, ~ and ~ y \le -1$. These two can be added to get $y-3 \le x-1 \implies x-y \ge -2.$ So finally we have $$ -2\le x-y \le 7$$