Question

How do you solve `\frac { 2- 2x } { 3x + 5} \leq 0`?

Answer 1

Solution is `x <= -5/3` or `x >=1`

Explanation:

For considering `(2-2x)/(3x+5) <=0`, let us first consider `(2-2x)/(3x+5) <0`

If it is so either `2-2x <0` and `3x+5 > 0` i.e. `x>1` or `x> -5/3` i.e. `x > 1`

or `2-2x >0` and `3x+5 < 0` i.e. `x<1` or `x< -5/3` i.e. `x< -5/3`

Now adding, solution is `x <= -5/3` or `x >=1`

Answer 2

The solution is `x in (-oo,-5/3) uu[1, +oo)`

Explanation:

Let `f(x)=(2(1-x))/(3x+5)`

We build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaa)` `-oo` `color(white)(aaaaaa)` `-5/3` `color(white)(aaaaaa)` `1` `color(white)(aaaaaaa)` `+oo`

`color(white)(aaaa)` `3+5x` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `+` `color(white)(aa)` `0` `color(white)(aaa)` `+`

`color(white)(aaaa)` `1-x` `color(white)(aaaaaaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aa)` `+` `color(white)(aa)` `0` `color(white)(aaa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `+` `color(white)(aa)` `0` `color(white)(aaa)` `-`

Therefore,

`f(x)<=0` when `x in (-oo,-5/3) uu[1, +oo)`

graph{(2-2x)/(3x+5) [-14.24, 14.25, -7.12, 7.12]}