Question

P is partly constant and partly varies inversely as Q.lf Q=9 when P=3 and Q=18 when P=9 find P when Q=12

Answer 1

P = k/Q + C

given:
P=3, Q=9
3 = k/9 + C
27 = K + 9C ***

P=9, Q = 18
9 = k/18 + C
162 = k + 18C **

subtract *** from **
135 = 9C
C = 15
in ***
k+135 = 27
k = -108

P = -108/Q + 15

so when Q = 12
P = -108/12 + 15 = 6

Answer 2

P=-108\12+ 15 =6
Multiply both side /no. With 12.
6 = 15 -108\12, you get
72 = 180 - 108
72 = 72, cross over 72 to the other side.
72-72 = 0

So, P = 0