Question

How do you solve `(4-2x)/(3x+4)<=0`?

Answer 1

The answer is `=x in ] -4/3,2]`

Explanation:

Let `f(x)=(4-2x)/(3x+4)`

The domain of `f(x)` is `D_f(x)=RR-{-4/3}`

Now,

we construct a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaa)` `-4/3` `color(white)(aaaaaa)` `2` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `3+4x` `color(white)(aaaaa)` `-` `color(white)(aa)` `color(red)(∥)` `color(white)(aa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `4-2x` `color(white)(aaaaa)` `-` `color(white)(aa)` `color(red)(∥)` `color(white)(aa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `+` `color(white)(aa)` `color(red)(∥)` `color(white)(aa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(x)<=0`, when `x in ] -4/3,2]`