Question #c6156

1 Answer
May 28, 2017

Answer:

The solution is #x in (2,7/3) uu (3.+oo)#

Explanation:

#x^2-5x+6=(x-2)(x-3)#

Let's simplify the inequality, without crosing over

#3/(x^2-5x+6)+(4-x)/(3-x)>(6-x)/(2-x)#

#3/((x-2)(x-3))-(4-x)/(x-3)+(6-x)/(x-2)>0#

#(3-(4-x)(x-2)+(6-x)(x-3))/((x-2)(x-3))>0#

#(3-4x+8+x^2-2x+6x-18-x^2+3x)/((x-2)(x-3))>0#

#(3x-7)/((x-2)(x-3))>0#

Let #f(x)=(3x-7)/((x-2)(x-3))#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaa)##2##color(white)(aaaaaa)##7/3##color(white)(aaaaaaa)##3##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x-2##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aa)##+#

#color(white)(aaaa)##3x-7##color(white)(aaa)##-##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aa)##+#

#color(white)(aaaa)##x-3##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+#

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

Therefore,

#f(x)>0#, when #x in (2,7/3) uu (3.+oo)#