I know how to add or subtract two inequalities. But I cannot find in anywhere how to add or subtract two compound inequalities. This problem is one of the example :
If $0 < x \leq 2$, and $y \geq 2$, determine $y-x$ and $x+y$.
This is my first time using stack exchange and I don't know how to use MathJax formatting. I appreciate your help so much.
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You can also work with intervals: $$\begin{align}0 < x \leq 2 \Rightarrow &x\in (0,2]\\ y\ge 2 \Rightarrow &y\in [2,+\infty)\end{align}$$ For $x+y$, you add the left (min) and right (max) borders (preserving the stricter inequality sign if any): $$\quad \quad \quad \quad x+y\in (2,+\infty)$$ You can multiply an interval by constants (if negative, the interval is reversed): $$\begin{align}x\in (0,2] \Rightarrow -x&\in [-2,0)\\ y&\in [2,+\infty)\end{align}$$ For $y-x=y+(-x)=-x+y$, you again add the left (min) and right (max) borders (preserving the stricter inequality sign if any): $$-x+y=y-x\in [0,+\infty).$$ Exercise: Can you find $x-y$ ?
Adding inequalities is fairly straightforward, so you can reason through a problem like this is this way: We have $0<x\le 2$, and we have $2\le y<\infty$. Therefore, $0+2<x+y<2+\infty$, which simplifies to: $2<x+y<\infty$.
To work with $y-x$, think of it as $y+(-x)$. We can find the inequalities for $-x$ by multiplying the inequalities for $x$ by $-1$: $-2\le -x<0$. This time, when we add, the left sides of both compound inequalities have $\le$ instead of $<$, so their sum does, as well: $2+(-2)\le y-x <0+\infty$.
Does that make sense?