4x<2x+1<//<3x+2
I simplify it down to 2x<1<//<1x+2
but I don't know where to go from there
The intequality is $\displaystyle 4x<2x+1\leq 3x+2$, a possible way to solve this is by considering three cases:
(1)$\displaystyle 4x<2x+1$
(2)$\displaystyle 4x\leq3x+2$
(3)$\displaystyle 2x+1\leq3x+2$
By solving this three inequality's and summarizing the solutions you will get the solutions for the original inequality.
You could subtract $\displaystyle 2x+1$ from all three terms.
$\displaystyle 4x-(2x+1)<2x+1-(2x+1)\le\ 3x+2-(2x+1)$
$\displaystyle 2x-1<0\le\ x+1$
$\displaystyle 2x-1<0$ is saying $\displaystyle 2x<1\Rightarrow\ x<\frac{1}{2}$
$\displaystyle 0\le\ x+1$ is saying $\displaystyle x\ge\ -1$
Then
$\displaystyle -1\le\ x<\frac{1}{2}$
I like this approach.
Solution sets:
$\displaystyle \begin{gathered} (1)\;A = \left( { - \infty ,\frac{1}{2}} \right) \hfill \\ (2)\;B = \left( { - \infty ,2} \right] \hfill \\ (3)\;C = \left[ { - 1,\infty } \right) \hfill \\ \end{gathered} $
So $\displaystyle A \cap B \cap C = \left[ { - 1,\frac{1}{2}} \right)$
Hello, GaDoomZ!
$\displaystyle 4x \:<\:2x+1 \:\le\: 3x+2$
Separate the inequalities:
. . $\displaystyle 4x \:<\:2x+1 \quad\Rightarrow\quad 2x \:<\:1 \quad\Rightarrow\quad x \:<\:\tfrac{1}{2}$
. . $\displaystyle 2x+1\:\le\:3x+2 \quad\Rightarrow\quad -x \:\le\:1 \quad\Rightarrow\quad x \:\ge\:-1$
Therefore: .$\displaystyle -1\:\le\:x\:<\:\tfrac{1}{2}$
. . $\displaystyle \begin{array}{cccccc}---\!\!\!\!\! & \bullet\!\!\!\!\! & ===\!\!\!\!\! & \circ\!\!\!\!\! & ---\!\!\!\!\! \\ & \;\:\text{-}1 && \quad\frac{1}{2} \end{array}$