How do you solve and write the following in interval notation: # (x + 5)/(x + 9) < 3#?

1 Answer
Feb 10, 2018

Answer:

#x in (-oo, -11) uu (-9, oo)#

Explanation:

Given:

#(x+5)/(x+9) < 3#

I would like first to note what you can't do, which is to simply multiply both sides by #(x+9)#. The problem is that if #x < -9# then #(x+9) < 0#, so you would have to reverse the inequality too, while if #x > -9# then you would leave the inequality as #<#. So you would need to split into cases - messy.

So what can we do?

Noting first that #x != -9#, we can multiply both sides of the inequality by #(x+9)^2# which is positive, to get:

#(x+5)(x+9) < 3(x+9)^2#

which multiplies out to:

#x^2+14x+45 < 3x^2+54x+243#

Then subtracting #x^2+14x+45# from both sides we get:

#0 < 2x^2+40x+198 = 2(x^2+20x+99) = 2(x+9)(x+11)#

The right hand side is a quadratic with positive leading coefficient, which intersects the #x# axis at #x=-11# and #x=-9#.

Therefore it is positive when:

#x < -11" "# or #" "x > -9#

That is:

#x in (-oo, -11) uu (-9, oo)#