How do you solve and write the following in interval notation: #(x-2)/(x-3)<=-6#?

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Answer 1 · 2017-08-04T16:35:56

The solution is #x in [20/7, 3)#

We cannot do crossing over

Let's rearrange the inequality

#(x-2)/(x-3)<=-6#

#(x-2)/(x-3)+6<=0#

#((x-2)+6(x-3))/(x-3)<=0#

#((x-2+6x-18))/(x-3)<=0#

#(7x-20)/(x-3)<=0#

Let #f(x)=(7x-20)/(x-3)#

Let's build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaaa)##20/7##color(white)(aaaaaaaaaa)##3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##7x-20##color(white)(aaaa)##-##color(white)(aaaa)##0##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##+#

#color(white)(aaaa)##x-3##color(white)(aaaaaa)##-##color(white)(aaaa)##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aaaa)##0##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+#

Therefore,

#f(x)<=0# when #x in [20/7, 3)#

graph{(x-2)/(x-3)+6 [-32.47, 32.47, -16.24, 16.25]}