Question

The gradient of the curve y-xy+2px+3qy=0 at the point (3,2) is -2/3.the value of p and q are

Answer 1

write naa

dy/dx=d(xy)/dx+d(2px)/dx+d(3qy)/dx d(xy)/dx=xdy/dx+y 
dy/dx=xdy/dx+2p+3qdy/dx 
dy/dx(1-x-3q)=2p 
dy/dx=2p/(1-x-3q) 
substitute values: 
-2/3=2p/(1-3-3q) 
4+6q=6p ---->eq1 

y=xy+2px+3qy 
subs values: 
2=6+6p+6q 
6p+6q= -4 -->eq2 
Solve: 
subs eq1 into eq2 
6q+4+6q= -4 
12q= -8 
q= -2/3 
p=6q+4 
p=-4+4 
p=0 
Ans: p=0 q= -2/3

Answer 2

Step-by-step explanation:

therefore value of p=1\2,q=1\6