Graph D
Open holes at 1 and 3; shading in between
Explanation:
Let p = x-1 and q = x-3
The problem we want to solve is in the form p/q < 0
Since p/q is negative, there are two cases we have to address
Case 1) p is positive and q is negative, or,Case 2) p is negative and q is positive
The signs of p and q are opposite one another. Recall this is due to the fact that dividing a positive over a negative (or vice versa) leads to a negative value.
Let's go over Case 1
If p is positive, then
p > 0
x-1 > 0
x > 1
If q is negative, then
q < 0
x-3 < 0
x < 3
Combine x > 1 with x < 3 and we get 1 < x < 3. We'll come back to this later.
Now case 2
If p is negative, then
p < 0
x-1 < 0
x < 1
If q is positive, then,
q > 0
x-3 > 0
x > 3
We get a contradiction.
Both "p is negative" AND "q is positive" must be true at the same time. But if p is negative, then x < 1 which contradicts x > 3 (which happens when q is positive)
Put simply: case 2 is not possible. There's no way to have p be negative at the same time q is positive. This because we're wanting x to both be smaller than 1 AND also be greater than 3 at the same time.
We must rule out case 2 and side with case 1 instead.
The conclusion of case 1 was that 1 < x < 3
The graph of this has open holes at 1 and 3 on the number line. The shading is between the open holes to represent all values between 1 and 3, excluding the endpoints. Graph D indicates these features, so that's why graph D is the answer.
Note: the open holes indicate that x = 1 and x = 3 are not part of the solution set. Another reason x = 3 is restricted is to avoid a division by zero error.
