In a remote school of 13 teachers, 8 teach Government (G), 7 teach Geography (P) and 7 teach History (H). 2 teach Government and Geography only, 2 teach Government and History only and 3 teach Geography and History only. Represent the information in a Venn diagram.

here's how to solve this problem involving sets and Venn diagrams: First, let's define the sets and the given information: Let N be the total number of teachers in the school. N = 13. Let G be the set of teachers who teach Government. n(G) = 8. Let P be the set of teachers who teach Geography. n(P) = 7. Let H be the set of teachers who teach History. n(H) = 7. We are given the following "only" intersections: n(G P only) = 2 (teachers who teach Government and Geography but not History) n(G H only) = 2 (teachers who teach Government and History but not Geography) n(P H only) = 3 (teachers who teach Geography and History but not Government) Step 1: Find the number of teachers who teach all three subjects, n(G P H). Let x = n(G P H). The number of teachers who teach only Government is n(G only) = n(G) - [n(G P only) + n(G H only) + x]. n(G only) = 8 - (2 + 2 + x) = 8 - (4 + x) = 4 - x. The number of teachers who teach only Geography is n(P only) = n(P) - [n(G P only) + n(P H only) + x]. n(P only) = 7 - (2 + 3 + x) = 7 - (5 + x) = 2 - x. The number of teachers who teach only History is n(H only) = n(H) - [n(G H only) + n(P H only) + x]. n(H only) = 7 - (2 + 3 + x) = 7 - (5 + x) = 2 - x. The sum of all disjoint regions must equal the total number of teachers, assuming all teachers teach at least one subject. n(G P H) = n(G only) + n(P only) + n(H only) + n(G P only) + n(G H only) + n(P H only) + n(G P H) 13 = (4 - x) + (2 - x) + (2 - x) + 2 + 2 + 3 + x 13 = 4 - x + 2 - x + 2 - x + 2 + 2 + 3 + x ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._