Equality

[mathematics]

I am in need of this help pleaseee

Determine the solution region of ; 3x(2y-3)<0

Please show me how to do this

 

[soroban]

Hello, mathematics!

Determine the solution region of: .$\displaystyle 3x(2y-3)\:<\:0$

$\displaystyle \text{The product of }3x\text{ and }2y-3\text{ is negative.}$


$\displaystyle \text{There are two cases:}$

. . $\displaystyle [1]\;3x \,>\,0\:\text{ and }\:2y-3\:<\:0$
. . $\displaystyle [2]\;3x\,<\,0\:\text{ and }\:2y-3\,>\,0$


$\displaystyle \text{Case [1]: }\;\begin{array}{ccccccc}3x \:>\: 0 & \Rightarrow & x \:>\: 0 \\2y-3 \:<\: 0 & \Rightarrow & y \: < \: \frac{3}{2} \end{array}$


$\displaystyle \text{Case [2]: }\;\begin{array}{cccc}3x\:<\:0 & \Rightarrow & x \:<\:0 \\ 2y-3 \:>\:0 & \Rightarrow & y \:>\:\frac{3}{2} \end{array}$


$\displaystyle \text{The solution region looks like this:}$

           :::::|
          ::::::|
         :::::::|
        ::::::::|
      - - - - - + - - - - - -  y = 3/2
                |:::::::::
                |::::::::
      ----------+:-:-:-:-------
               0|::::::
                |:::::
                |::::