How do you solve #(3x+5)/(6-2x)<=0#?

1 Answer
Jan 19, 2017

Answer:

The answer is #x in ] -oo,-5/3 ] uu ] 3, +oo [ #

Explanation:

Let #f(x)=(3x+5)/(6-2x)#

The domain of #f(x)# is #D_f(x)=RR-{3}#

Now, we can make the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-5/3##color(white)(aaaaaaa)##3##color(white)(aaaaaaaaaaa)##+oo#

#color(white)(aaaa)##3x+5##color(white)(aaaa)##-##color(white)(aaaaaa)##+##color(white)(aa)##∥##color(white)(aaaa)##+#

#color(white)(aaaa)##6-2x##color(white)(aaaa)##+##color(white)(aaaaaa)##+##color(white)(aa)##∥##color(white)(aaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaaaa)##+##color(white)(aa)##∥##color(white)(aaaa)##-#

Therefore,

#f(x)<=0#, when #x in ] -oo,-5/3 ] uu ] 3, +oo [ #