Question

How do you solve `(x+3)/(3x-6)<=2`?

Answer 1

`x in (-oo, 2) uu [3, oo)`

Explanation:

As per the question, we have

`(x+3)/(3x-6) <= 2`

`:. (x+3)/(3x-6) - 2 <= 0`

`:. (x + 3 -6x + 12)/(3x - 6) <= 0`

`:. (-5x + 15)/(3x - 6) <= 0`

`:. (5x - 15)/(3x - 6) >= 0`

`:. (5(x-3))/(3(x-2)) >= 0`

`:.` By Wavy Curve Method we get,

Note : In the image the orange region represents the area on the numberline to which `x` belongs, i.e from `-oo` to `2` and `3` to `oo`.

`x in (-oo, 2) uu [3, oo)`

Hence, the answer.

Answer 2

`color(blue)((-oo,2)uu[3,oo)`

Explanation:

`(x+3)/(3x-6) <=2`

Multiply by `(3x-6)^2` ( valid since this is always positive )

`(3x-6)(x+3)<=2(3x-6)^2`

`3x^2+3x-18<=18x^2-72x+72`

`15x^2-75x+90>=0`

Factor:

`(15x-30)(x-3)>=0`

`(15x-30)(x-3)=0=>x=2,x=3`

Using signs of brackets:

For:

`2<=x<=3`

`+ - >=0color(white)(888)` False

For:

`x<=2`

`- - >=0color(white)(888)` True

For:

`x>=3`

`+ + >=0color(white)(888)` True

Solutions in interval notation:

`(-oo,2)uu[3,oo)`

Notice the use of a open interval for 2, this is because for `x= 2` the denominator would be zero.