# Demand Equation Question

• September 10th 2007, 04:36 PM
redraider717
Demand Equation Question
Suppose that 10,000 units of a certain item are sold per day by the entire industry at a price of \$150 per item, and that 8000 units can be sold per day by the same industry at a price of \$200 per item. Find the demand equation for http://webwork.math.ttu.edu/webwork2...da050b4dd1.png in terms of http://webwork.math.ttu.edu/webwork2...dd0b8b8e91.png, the number of units sold, assuming the demand curve to be a straight line.

I know the answer will be a straight line, so P(x)=ax+b

and I know

if you sell 10000 units, price becomes \$150 so in price equation above you get 150=a(10000)+b

and

if you sell 8000 units, price becomes \$200 so the equation becomes 200=a(8000)+b

I just can't figure out how to solve to get A and B pretty much...
• September 10th 2007, 06:05 PM
Jhevon
Quote:

Originally Posted by redraider717
Suppose that 10,000 units of a certain item are sold per day by the entire industry at a price of \$150 per item, and that 8000 units can be sold per day by the same industry at a price of \$200 per item. Find the demand equation for http://webwork.math.ttu.edu/webwork2...da050b4dd1.png in terms of http://webwork.math.ttu.edu/webwork2...dd0b8b8e91.png, the number of units sold, assuming the demand curve to be a straight line.

I know the answer will be a straight line, so P(x)=ax+b

and I know

if you sell 10000 units, price becomes \$150 so in price equation above you get 150=a(10000)+b

and

if you sell 8000 units, price becomes \$200 so the equation becomes 200=a(8000)+b

I just can't figure out how to solve to get A and B pretty much...

these are simultaneous equations

$150 = 10000a + b$ ..................(1)
$200 = 8000a + b$ ....................(2)

$\Rightarrow -50 = 2000a$ ....................(1) - (2)

now solve for a. when finished, plug the value you got for a into either of the original equations to solve for b