In a survey of 100 outpatients who reported at a hospital one day, it was found that 70 complained of fever, 50 had stomach trouble, and 30 were injured. all the 100 patientshad one or the other of these complaints, and 44 had exactly two of them. how many patients had all three complaints?

6 patients had all three complaints.
We can't subtract 44 from 30.
we can subtract that from from 50 which leaves us with 6 patients that can have the third complaint as well.
We can also subtract 44 from 70 but the remaining number exceeds the number of patients with stomach ache.
so it leaves us with 6.
Revising:
So according to the survey
All 44 have exactly two complaints
30 patients were injured
50 had stomach ache
70 had fever
1. 30 is less than 44 so that leaves it outside
2. 50 is greater than 44 so we subtract 44 from 50. 6 patients are those who can have all three complaints.
3. 70 is also greater than 44 so we subtract 44 from 70. 26 patients are left behind.
The number of patients can't be higher than 6 because that is the number of patients that have stomach ache and can also have the other two complaints. Hence 6 patients have all three complaints.

the formula of a slope:

we have the points (-2, 7) and (2, 3). substitute:

answer: the slope is equal -1.