1. 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

2. Re: Solving for X

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
$p = -0.02x + 400$

note that $0.02 = \frac{2}{100} = \frac{1}{50}$ ...

$p = -\frac{1}{50} x + 400$

$p - 400 = -\frac{1}{50} x$

$-50p + 20000 = x$

3. Re: Solving for X

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