How many of the integers that satisfy the inequality (x+2)(x+3) / x-2 ≥ 0 are less than 5?

A) 1

B) 2

C) 3

D) 4

E) 5

One approach is to determine the CRITICAL POINTS: the values where the lefthand side is EQUAL TO 0 or is UNDEFINED.
The lefthand side is equal to 0 when x=-2 and x=-3.
The lefthand side is undefined when x=2.

We already know that x=-2 and x=-3 are valid solutions because they are where (x+2)(x+3) / x-2 = 0.
To determine the range where (x+2)(x+3) / x-2 > 0, try one integer value to the left and right of each critical point.

x < -3:
Plugging x=-4 into (x+2)(x+3) / x-2 > 0, we get:
(-4+2)(-4+3)/(-4-2) > 0
2/-6 > 0.
Doesn't work.
This means that no value less than -3 will work.

-3<x<-2:
No integer values in this range.

-2<x<2:
Plugging x=0 into (x+2)(x+3) / x-2 > 0, we get:
(0+2)(0+3)/(0-2) > 0
-3 > 0.
Doesn't work.
This means that no value between -2 and 2 will work.

x>2:
Plugging x=3 into (x+2)(x+3) / x-2 > 0, we get:
(3+2)(3+3)/(3-2) > 0
30 > 0.
This works.
This means that ANY VALUE greater than 2 will work.
There are only two integer values between 2 and 5:
3 and 4.

Thus, there are four integer values less than 5 that satisfy the inequality: -3, -2, 3 and 4.

The correct answer is D.

Other problems that I've solved with the critical point approach:
https://www.beatthegmat.com/inequality-c ... 89518.html
https://www.beatthegmat.com/knewton-q-t89317.html
https://www.beatthegmat.com/which-is-true-t89111.html=

Consider the equation
(x+2)(x+3)/x - 2 = 0
(x+2)(x+3)- 2x = 0
x^2 + 5x + 6 - 2x = 0
y = x^2 + 3x + 6 = 0
This is a U-curve.

dy/dx = 2x + 3 = 0 gives a minimum at x = -3/2

Therefore the lowest possible integer is -1 and the highest integer is 4 (from the question the integers are less than 5).
This gives us 6 values: -1, 0, 1, 2, 3, 4
However, y >= 0 and x = -1, 0 both yield y < 0
So the answer is D

Here's an easier way.

We know that x ≠ 2, or we're dividing by 0.

Let's assume x > 2. This only gives us two cases to check: x = 3 and x = 4. These both work, so we're set there.

Now let's assume x < 2. That means (x - 2) is negative. Now we'll multiply both sides by (x - 2) to get rid of the denominator. Since we multiplied by a negative, the sign FLIPS, giving us

(x + 2)(x + 3) ≤ 0

This will EQUAL 0 if (x+2) = 0 or if (x+3) = 0, so x = -2 and x = -3 are also solutions.

(x + 2)(x + 3) < 0 if and only if (x + 2) is negative and (x + 3) is positive. But this only works for -3 < x < -2, so there are no integer solutions.

Hence we have four solutions: -3, -2, 3, and 4.