How do you solve #x/(3-x)>2/(x+5)# using a sign chart?

1 Answer
Feb 1, 2017

Answer:

The answer is #x in ]-7.77, -5[uu ]0.77,3[#

Explanation:

We cannot do crossing over

We rewrite the equation

#(x)/(3-x)>(2)/(x+5)#

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

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

#(x^2+5x-6+2x)/((3-x)(x+5))=(x^2+7x-6)/((3-x)(x+5))#

Let #f(x)=(x^2+7x-6)/((3-x)(x+5))#

The roots of #x^2+7x-6# are

#x=(-7+-sqrt(49+4*6))/2#

#x=(-7+-sqrt73)/2#

#x_1=(-7+8.54)/2=0.77#

#x_2=(-7-8.54)/2=-7.77#

Now, we can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaa)##-7.77##color(white)(aaaa)##-5##color(white)(aaaa)##0.77##color(white)(aaaa)##3##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x+7.77##color(white)(aaaaa)##-##color(white)(aaaaaa)##+##color(white)(aa)##||##color(white)(aa)##+##color(white)(aaa)##+##color(white)(aa)###||###color(white)(aaa)##+#

#color(white)(aaaa)##x+5##color(white)(aaaaaaaa)##-##color(white)(aaaaaa)##-##color(white)(aa)##||##color(white)(aa)##+##color(white)(aaa)##+##color(white)(aa)###||###color(white)(aaa)##+#

#color(white)(aaaa)##x-0.77##color(white)(aaaaaa)##-##color(white)(aaaaaa)##-##color(white)(aa)##||##color(white)(aa)##-##color(white)(aaa)##+##color(white)(aa)###||###color(white)(aaa)##+#

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

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaaa)##-##color(white)(aaaaaa)##+##color(white)(aa)##||##color(white)(aa)##-##color(white)(aaa)##+##color(white)(aa)###||###color(white)(aaa)##-#

Therefore,

#f(x)>0# when #x in ]-7.77, -5[uu ]0.77,3[#