How do you solve #x/(x-5) >= -2 #?

Question

1968 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-24T17:22:15

#x>=10/3#

#x/(x-5) >=-2#
#x >= -2(x-5)#
#x>= -2x + 10#
#3x>=10#
#x>=10/3#

Answer 2 · 2018-03-24T17:22:25

#x>=10/3#

#x>=-2(x-5)# multiply both sides by #x-5# to get rid of fraction
#x>=-2x+10# distribute
#3x>=10# add 2x to both sides
#x>=10/3#

Answer 3 · 2018-03-24T17:23:26

#x<=10/3 or x>5#

#x/(x-5)>=-2#

We wanna find the critical points of the inequality

#x/(x-5)=-2#

Cross multiply

#x = (-2)(x) + (-2)(-5)#

#x= -2x + 10#

Add #2x# on both sides

#x + 2x = -2x + 10 + 2x#

#3x = 10#

Then divide both sides by #3#

#(cancel3x)/(cancel3)=10/3#

#x=10/3#

Don't forget we were looking for the critical points:

Critical points

#x=10/3# which makes both sides equal

#x=5#

Ok now we can check the intervals in between critical points.

We have #x<= 10/3# which works in the original inequality

We also have #10/3 <= x <5# which doesn't work in the original
inequality

And we have #x > 5# which works in the original inequality

Thus,

The answers are:

#x <= 10/3 or x > 5#