# What is the solution of `3/(x+5)lt=6` ?The slated lines are suppose to be straight. also the > is underlined

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`3/(x+5)lt=6`

To solve, set the right side equal to zero. So, subtract both sides by 6.

`3/(x+5) - 6 lt = 6 - 6`

`3/(x+5) - 6 lt=0`

Then, express left side as one fraction.

`3/(x+5) - (6(x+5))/(x+5)lt=0`

`3/(x+5) -(6x+30)/(x+5)lt=0`

`(-6x-27)/(x+5)lt=0`

Then, set the numerator and the denominator equal to zero. And, solve for x.

`-6x-27=0 ` and `x + 5=0`

`-6x=27 ` `x=-5`

`x=-27/6`

`x=-4.5`

These values of x are the critical numbers. Note that the critical numbers refer to the boundaries of the interval. Since there are two critical numbers, there are three possible intervals that satisfy the inequality equation. But we have to consider that we cannot as -5 as a value of x since it would give us a zero denominator.

So the possible solutions are `xlt-5` , `-5ltxlt=-4.5` and `xgt-4.5` .

From here, we need need to assign a test point to determine which among them gives us a true condition.

For the interval `xlt-5` , let the test point be `x=-6` .

`3/(x+5)lt=6`

`3/(-6+5)lt=6`

`-3lt=6` (True)

This means that the interval `xlt-5 ` is a solution to the inequality equation.

For the interval; -`5ltxlt=-4` .5, let `x=-4.8` .

`3/(-4.8+5)lt=6`

`15lt=6 ` (False)

Hence, `-5ltxlt=-4.5 ` is not a solution.

And for the interval `xgt=-4.5` , let `x=0` .

`3/(0+5)lt=6`

`3/5lt=6` (True)

Thus, `xgt=-4.5` is a solution.

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**Answer: The solution to the given inequality equation is `xlt-5` and ` xgt=-4.5` . In interval notation, we may express this as `(-oo,-5) ` `uu` `[-4.5,+oo)` .**

`3/(x+5) <= 6`

`3 <= 6(x+5)`

`3 <= 6x+30`

`3-30 <= 6x`

`-27 <= 6x`

`-27/6 <= x`

`-4.5 <= x`

**So all the x values that is less than of equal to -4.5 satisfies the above inequality.**

`x in (-oo,-4.5]`