A nursery sells Kentucky Blue Grass seed for $5.75 per pound and Tall Fescue seed for$4.50 per pound. The nursery sells a mixture of the two kinds of seed for$5.25 per pound. Let k represent the amount of Kentucky Blue Grass seed the nursery uses in 5 pounds of the mixture. Write an equation to represent the problem. How much Kentucky Blue Grass does the nursery use in 5 pounds of the mixture? How much Tall Fescue does the nursery use in 5 pounds of the mixture?

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To solve this problem, let's break it down step by step:

Step 1: Define the variables:
Let k represent the amount of Kentucky Blue Grass seed the nursery uses in 5 pounds of the mixture.
Let f represent the amount of Tall Fescue seed the nursery uses in 5 pounds of the mixture.

Step 2: Determine the given information:
- The price of Kentucky Blue Grass seed is $5.75 per pound.
- The price of Tall Fescue seed is $4.50 per pound.
- The price of the mixture is $5.25 per pound.
- We want to find the amount of Kentucky Blue Grass seed (k) used in 5 pounds of the mixture.
- We also want to find the amount of Tall Fescue seed (f) used in 5 pounds of the mixture.

Step 3: Set up the equation:
The equation to represent the problem can be derived from the fact that the total price of the mixture must equal the sum of the prices of the individual components.

Since we have the prices per pound, the equation becomes:

(Price of Kentucky Blue Grass seed per pound * Amount of Kentucky Blue Grass seed used) +
(Price of Tall Fescue seed per pound * Amount of Tall Fescue seed used) =
Price of the mixture per pound * Total weight of the mixture.

In this case, the equation is:
(5.75k) + (4.50f) = 5.25 * 5

Step 4: Solve for k:
To find the amount of Kentucky Blue Grass seed used in 5 pounds of the mixture (k), we will isolate k in the equation.

5.75k + 4.50f = 26.25

Step 5: Substitute values and solve:
Since the equation is for 5 pounds of the mixture, we know that the total weight is 5 pounds.

Let's substitute this value into the equation:
5.75k + 4.50f = 26.25

Now, we can solve for k. However, since the equation contains two unknowns (k and f), we cannot find a unique value for k without additional information.