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

2 Answers

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

Jul 11, 2017

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]