How do you solve \frac{- x - 3}{x + 2} \leq 0x3x+20?

2 Answers

HI
given equation,
(-x-3)/(x+2)x3x+2<=0

thus ,to solve this question we need to get the denominator term (x+2)(x+2) in numerator,
thus
=(-x-2-1)/(x+2)x21x+2
=(-x-2)/(x+2)x2x+2 -1/(x+2)1x+2,
now,
getting minus sign common,
from first term,
=(-(x+2)/(x+2))(x+2x+2)-1/(x+2)1x+2
=-1-1/(x+2)1x+2
as this term is less than and equal to zero
hence,
-1<=(1/(x+2))(1x+2)
now,getting (x+2) to the numerator side to left and minus one to right
(x+2)<=-1(x+2)1
x<=-1-2x12
x<=-3x3,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 > -2x3orx>2, in interval notation (-oo,-3]uu (-2,oo)(,3](2,)

Explanation:

(-x-3) / (x+2) <= 0 ; x != -2 x3x+20;x2

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]