I tried to understand what you wrote
1+2/3 -1 1/6 I got 1/2
The answer is both all real numbers less than 0 and all real numbers greater than 0
60 Student
5/7=0.71428571428<span>
</span>60/84=0.71428571428
We are given that there are a total of 78 students. If we set the following variables:
![\begin{gathered} C=\text{students only in chemestry} \\ P=\text{students only in physics} \\ PC=\text{students in physics and chemistry} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20C%3D%5Ctext%7Bstudents%20only%20in%20chemestry%7D%20%5C%5C%20P%3D%5Ctext%7Bstudents%20only%20in%20physics%7D%20%5C%5C%20PC%3D%5Ctext%7Bstudents%20in%20physics%20and%20chemistry%7D%20%5Cend%7Bgathered%7D)
Then, the sum of all of these must be 78, that is:
![C+P+PC=78](https://tex.z-dn.net/?f=C%2BP%2BPC%3D78)
Since there are 15 in chemistry and physics and 47 in chemistry, we may replace that into the equation and we get:
![47+P+15=78](https://tex.z-dn.net/?f=47%2BP%2B15%3D78)
Simplifying:
![62+P=78](https://tex.z-dn.net/?f=62%2BP%3D78)
Now we solve for P by subtracting 62 on both sides:
![\begin{gathered} 62-62+P=78-62 \\ P=16 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%2062-62%2BP%3D78-62%20%5C%5C%20P%3D16%20%5Cend%7Bgathered%7D)
Therefore, there are 16 students in physics