Question #14e0c

Question

908 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-27T07:00:55

The solution is #x in (-oo,-3) uu [-8/7,+oo)#

Let 's rewrite and simplify the equation, we cannot do crossing over

#(7-2x)/(x+3)<=5#

#5-(7-2x)/(x+3)>=0#

#(5(x+3)-(7-2x))/(x+3)>=0#

#(5x+15-7+2x)/(x+3)>=0#

#(7x+8)/(x+3)>=0#

Let #f(x)=(7x+8)/(x+3)#

We build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaaa)##-3##color(white)(aaaaaa)##-8/7##color(white)(aaaa)##+oo#

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

#color(white)(aaaa)##7x+8##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##-##color(white)(aaaaa)##+#

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

Therefore,

#f(x)>=0# when #x in (-oo,-3) uu [-8/7,+oo)#