Solve Compound Inequality Without Separation With Variables on Outside I have the following compound inequality:
$3y-5 < -2y < 2+y$
I know I can separate them into two different equations, and go from there. I have been taught how to solve them without separation when the variable is in the center. But I have not been taught how to do this when there are variables on the outside like this one? How do you I "put the y variable back into the center?
The way I see it, I  can't do it. I can -y from both sides and add it to the center, but still leaves a 2y on the left. I can -3y and add it to the center, but that leaves a -2y on the right. The only way to solve it is through separation. Or am I wrong?
 A: Compound inequalities are just single inequalities that must be simultaneously true.  In other words, $$3y - 5 < -2y < 2 + y$$ is just a more compact way of saying $$3y - 5 < -2y \quad {\it and} \quad -2y < 2 + y.$$  Then you solve the individual inequalities:
$$3y - 5 < -2y$$ implies $$5y < 5$$ which is equivalent to $$y < 1.$$  The second inequality is $$3y > -2$$ or $$y > -\frac{2}{3}.$$  Now you put them back together; we must simultaneously have $y < 1$ and $y > -\frac{2}{3}$, or $$-\frac{2}{3} < y < 1.$$
I believe this process is what you call "separation."  I do not advise trying any other shortcut method that involves working both inequalities at the same time.  It's not that it can't be done, it's that it's potentially confusing and you may make a mistake, especially as the inequalities become more complicated.  Even in this case, you can see that there is no way to algebraically add or subtract any multiple of $y$ to both inequalities simultaneously in order to eliminate $y$ from the extremes.  This is because the the LHS inequality $3y - 5 < -2y$ leads to the RHS inequality $y < 1$, and vice versa.  The "ends" become the "middle."
A: $$3y-5 < -2y < 2+y\implies 2y-5 \lt -3y \lt 2$$
$$2y-5 \lt -3y\implies 5y\lt 5\implies y \lt 1 
\quad \land \quad  
-3y \lt 2\implies -2 \lt 3y\implies -\frac{2}{3} \lt y$$
$$\therefore -\frac{2}{3} \lt y \lt 1$$
