Thread: Solving for X

Q:Solving the given demand equation for x in terms of p,

p= -0.02x+400 (o< or = x < or =20000)

text book answer: x=f(p) = -50p + 20000

I got x = -p/0.02 + 20000

I don't understand how is it 50p.

Thanks

Originally Posted by hovermet

Q:Solving the given demand equation for x in terms of p,

p= -0.02x+400 (o< or = x < or =20000)

text book answer: x=f(p) = -50p + 20000

I got x = -p/0.02 + 20000

I don't understand how is it 50p.

Thanks

note that ...

Originally Posted by hovermet

Q:Solving the given demand equation for x in terms of p,

p= -0.02x+400 (o< or = x < or =20000)

text book answer: x=f(p) = -50p + 20000

I got x = -p/0.02 + 20000

I don't understand how is it 50p.

Thanks

1/(0.02) = 50