Question

How do you solve `x- \frac { 1} { 2- 3x } > \frac { 2x - 1} { 2} + \frac { 6x + 1} { 3x - 2}?`

Answer 1

The solution is `x in (-2/9,2/3)`

Explanation:

Let' simplify the inequality

`x-1/(2-3x)>(2x-1)/2+(6x+1)/(3x-2)`

`x-1/(2-3x)-(2x-1)/2-(6x+1)/(3x-2)>0`

`x-1/(2-3x)-(2x-1)/2+(6x+1)/(2-3x)>0`

`(2x(2-3x)-2-(2x-1)(2-3x)+2(6x+1))/(2(2-3x))>0`

`((2x-3)(2x-2x+1)-2+2(6x+1))/(2(2-3x))>0`

`(2-3x+12x+2-2)/(2(2-3x))>0`

`(2+9x)/(2(2-3x))>0`

Let `f(x)=(2+9x)/(2(2-3x))`

We can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-2/9` `color(white)(aaaaaa)` `2/3` `color(white)(aaaaaaa)` `+oo`

`color(white)(aaaa)` `2+9x` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `2-3x` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aaaa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aaaa)` `-`

Therefore,

`f(x)>0` when `x in (-2/9,2/3)`