How is it possible to solve for the possible values of $x$ given the following information?
$\frac{a}{b}=\frac{2}{3},$ where $a \in [x, x+1)$ and $b \in [x,x+\frac{1}{2})$.
I can see that a valid value of $x$ would be $0$, and that $10$, for example, would be invalid, but I'm unsure how to find and prove a set of values.
$a \in [x,x+1)$ and $b =\frac {3a} 2 \in [x, x+\frac 1 2)$ so $ a\in [\frac {2x} 3, \frac {2x+1} 3)$. This is possible iff $[x,x+1)$ and $[\frac {2x} 3, \frac {2x+1} 3)$ have a point in common which means neither interval is completely to the left of the other. The conditions for this are $x+1 >\frac {2x} 3$ and $\frac {2x+1} 3 >x$ which translates to $-3 <x<1$.
We have $3a = 2b$, so we require $$ 3[x,x+1) \cap 2[x,x+1/2) \not = \varnothing \text{,} $$ or what is the same thing $$ [3x,3x+3) \cap [2x,2x+1) \not = \varnothing \text{.} $$ This intersection is empty if $3x+3 \leq 2x$ or $2x+1 \leq 3x$. The first case is $x \leq -3$ and the second case is $1 \leq x$. Recall that either of these is a condition for the intersection to be empty, so we want both of these to be false, so we want $x > -3$ and $1 > x$. Therefore, the complete set of values of $x$ which permit the given equation is $x \in (-3,1)$.
