I was working with inequalities and came up with a problem that could be simplified to something like this:
If $1 < x < 3\, $
Then,
$\,2 < 2x < 6 \,$ and $\, -9 < -3x < -3 $
Adding the two inequalities together, we end up with:
$ -7 < -x < 3\,\,(1)$
But,
$-3 < -x < -1\,\, (2)$
So, I want to know, where did I mess up? Are both statments equal? If so, is there a way to came up with $(2)$ from $(1)$?
When you added the two equations together you lost information that you cannot recover from the resulting equation.
It's a little easier to explain this when the inequalities are non-strict. Suppose $1 \leq x \leq 3.$ Then $2 \leq 2x \leq 6$ and $−9 \leq −3x \leq −3.$
Now it is certainly true that these last two sets of inequalities imply $-7 \leq -x \leq 3,$ because if $2 \leq y \leq 6$ and $−9 \leq z \leq −3$ then $-7 \leq y+z \leq 3.$ Just take $y = 2x$ and $z = -3x.$
But let's see what has to happen to achieve the lower bound, $-x = -7.$ Given $2 \leq y \leq 6$ and $−9 \leq z \leq −3,$ by referring to the inequalities before taking the sum we can see that the only way to achieve $y+z = -7$ is if $y = 2$ and $z = -9.$ But it is not possible that $2x = 2$ and $-3x = -9$ simultaneously.
It works just as poorly with strict inequalities, but we can only talk about numbers being close to the extreme values rather than at the extreme values so it is more complicated to explain.
To put it another way, after you reversed the direction of one of the sets of inequalities by multiplying by $-3,$ when you then added the inequalities you were simultaneously allowing the possibility that $x$ was near one end of its range $x \approx 1$ in one inequality and allowing the possibility that $x$ was near the other end of its range $x \approx 3$ in the other inequality when you combined the two inequalities to get a new lower or upper bound for $-x.$ You still get a true statement that way, but the statement is not as precise as it could be.
$$ -7 < -x < 3\kern.6em\not\kern -.6em \implies( 2 < 2x < 6 \:\text{ and } -9 < -3x < -3)$$ (counterexample: $x=5$);
therefore the step "Adding the two inequalities together, we end up with" has created the extraneous solutions $(-3,1]\cup[3,7).$
