A coffee merchant has coffee beans that sell for $9 per

pound and $12 per pound. The two types are to be mixed to create 100 lb of a mixture
that will sell for $11.25 per pound. How much of each type of bean should be
used in the mixture?

My problem is that I can't figure out the correct way to set this up.

This is how I am setting the problem up...then I am stuck.
900x+1200y=1125
-900(900x+1200)y=(-900)(1125)
-1,080,000y=-1,012,500
y=.9375

The final product will be 100 lbs selling for $1125. So 1125 = 9x + 12 (100-x). Where x is the number of pounds of the cheaper brand. You will then need (100 - x) pounds of the other brand to fill the mixture b/c x + (100-x) = 100. So solving 1125 = 9x + 12(100-x) leaves you with x = 25. So y (or 100-x) is 75.

To solve this problem correctly, you need to set up the equation in terms of the weight of the coffee beans, not the dollar amounts.

Let's say x represents the number of pounds of the $9 coffee beans, and y represents the number of pounds of the $12 coffee beans.

We are told that the total weight of the mixture is 100 pounds. So, we have the equation:

x + y = 100

We are also told that the average price per pound of the mixture is $11.25. Therefore, the total cost of the mixture can be calculated as follows:

(9x + 12y) / 100 = 11.25

Now, you have two equations:

x + y = 100 -- Equation 1
(9x + 12y) / 100 = 11.25 -- Equation 2

To solve this system of equations, you can use the method of substitution or elimination. I will demonstrate the method of substitution here.

From Equation 1, we can rewrite it as:

x = 100 - y

Now, substitute this expression for x in Equation 2:

(9(100 - y) + 12y) / 100 = 11.25

Simplifying:

(900 - 9y + 12y) / 100 = 11.25
900 + 3y = 1125
3y = 1125 - 900
3y = 225
y = 75

Now that we have the value of y, substitute it back into Equation 1 to find x:

x + 75 = 100
x = 100 - 75
x = 25

So, you need 25 pounds of the $9 coffee beans and 75 pounds of the $12 coffee beans to create a 100-pound mixture that sells for $11.25 per pound.