How do you solve #(4-2x)/(3x+4)<=0#?

1 Answer
Jan 10, 2017

Answer:

The answer is #=x in ] -4/3,2] #

Explanation:

Let #f(x)=(4-2x)/(3x+4)#

The domain of #f(x)# is #D_f(x)=RR-{-4/3}#

Now,

we construct a sign chart

#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaa)##-4/3##color(white)(aaaaaa)##2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##3+4x##color(white)(aaaaa)##-##color(white)(aa)##color(red)(∥)##color(white)(aa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##4-2x##color(white)(aaaaa)##-##color(white)(aa)##color(red)(∥)##color(white)(aa)##-##color(white)(aaaa)##+#

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

Therefore,

#f(x)<=0#, when #x in ] -4/3,2] #