Question

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

Answer 1

HI
given equation,
`(-x-3)/(x+2)`<=0

thus ,to solve this question we need to get the denominator term `(x+2)` in numerator,
thus
=`(-x-2-1)/(x+2)`
=`(-x-2)/(x+2)` `-1/(x+2)`,
now,
getting minus sign common,
from first term,
=`(-(x+2)/(x+2))` `-1/(x+2)`
=-1`-1/(x+2)`
as this term is less than and equal to zero
hence,
-1<=`(1/(x+2))`
now,getting (x+2) to the numerator side to left and minus one to right
`(x+2)<=-1`
`x<=-1-2`
`x<=-3`,upto infinity(negative)

And x cannot be equal to -2 as at x=-2 the given equation is not defied as denominator is 0
thus ,x can be greater than -2 upto positive infinity

Answer 2

Solution : `x <= -3 or x > -2`, in interval notation `(-oo,-3]uu (-2,oo)`

Explanation:

`(-x-3) / (x+2) <= 0 ; x != -2`

Crirical points are `-x-3=0 or x = -3 , x+2=0 or x = -2 :.`

Crirical points are `x = -3 , x = -2`

Sign change:
Interval ----------- Sign of `(-x-3) / (x+2)`

---

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

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

When `x > -2` ------ `(-)/(+) = (-) :. < 0`

When `x=-3 ; (-x-3) / (x+2)=0`

Solution : `x <= -3 or x > -2`, in interval notation `(-oo,-3]uu (-2,oo)`

graph{(-x-3)/(x+2) [-11.25, 11.25, -5.625, 5.62]} [Ans]