Find the possible values of x in (|x+3|+x)/(x+2)>1?

1 Answer
Jul 18, 2018

The solution is x in (-5,-2)uu(-1,oo)

Explanation:

There are 2 intervals to consider

I_1=(-oo,-3) and I_2=(-3,+oo)

In the first interval I_1

((-x-3)+x)/(x+2)-1>0

=>, (-3-(x+2))/(x+2)>0

=>, (-x-5)/(x+2)>0

let g(x)=(-x-5)/(x+2)

Build a sign chart

color(white)(aaaa)xcolor(white)(aaaaaa)-oocolor(white)(aaaa)-5color(white)(aaaa)-3

color(white)(aaaa)-x-5color(white)(aaaaa)+color(white)(aaaaa)-

color(white)(aaaa)x+2color(white)(aaaaaaa)-color(white)(aaaaa)-

color(white)(aaaa)g(x)color(white)(aaaaaaaa)-color(white)(aaaaa)+

Therefore, in the interval I_1

g(x)>0 when x in (-5,-3)

In the second interval I_2

((x+3)+x)/(x+2)-1>0

=>, (2x+3-(x+2))/(x+2)>0

=>, (x+1)/(x+2)>0

let h(x)=(x+1)/(x+2)

Build the sign chart

color(white)(aaaa)xcolor(white)(aaaa)-3color(white)(aaaaaa)-2color(white)(aaaaaa)-1color(white)(aaaa)+oo

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

color(white)(aaaa)x+1color(white)(aaaa)-color(white)(aaaa)#color(white)(aaaaa)-#color(white)(aaaa)+

color(white)(aaaa)h(x)color(white)(aaaaa)+color(white)(aaaa)||color(white)(aaaa)-color(white)(aaaa)+

Therefore, in this interval I_2

h(x)>0 when x in (-3,-2) uu (-1,+oo)

Finally,

The solution is

x in (-5,-3)uu(-3,-2) uu (-1,+oo)

That is

x in (-5,-2)uu(-1,oo)

graph{(|x+3|+x)/(x+2)-1 [-17.21, 11.27, -7.4, 6.84]}