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

2 Answers
May 9, 2018

Answer:

#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,

MS Paint
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.

May 9, 2018

Answer:

#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.