An auto manufacturer sends cars from two plants, I and II, to dealerships A and B located in a Midwestern city. Plant I has a total of 28 cars to send, and plant II has 8. Dealer A needs 20 cars, and dealer B needs 16. Transportation costs per car, based on the distance of each dealership form each plant, are $220 from I to A, $300 from I to B, $400 from II to A, $180 from II to B. The manufacturer wants to limit transportation costs to $10,640. How many cars should be sent from each plant to each of the two dealerships? (Assign variables to the 4 unknowns. Write a system of 5 equations with the 4 variables. Not all equations will involve all 4 variables)

No

Let's assign variables to the four unknowns:

Let's say the number of cars sent from Plant I to dealer A is x.
Let's say the number of cars sent from Plant I to dealer B is y.
Let's say the number of cars sent from Plant II to dealer A is z.
Let's say the number of cars sent from Plant II to dealer B is w.

Now, we can write a system of equations based on the given information:

Equation 1: x + y = 20 (Dealer A needs 20 cars)
Equation 2: z + w = 16 (Dealer B needs 16 cars)
Equation 3: x + z = 28 (Plant I has 28 cars)
Equation 4: y + w = 8 (Plant II has 8 cars)
Equation 5: 220x + 300y + 400z + 180w = 10,640 (Total transportation cost must be limited to $10,640)

Now, we can solve these equations to find the values of x, y, z, and w.

First, let's solve Equation 1 and Equation 3 simultaneously:

From Equation 1: x + y = 20
From Equation 3: x + z = 28

By subtracting Equation 1 from Equation 3, we get:

(z - y) = 8

Now, let's solve Equation 2 and Equation 4 simultaneously:

From Equation 2: z + w = 16
From Equation 4: y + w = 8

By subtracting Equation 4 from Equation 2, we get:

(z - y) = 8

We can see that both z - y = 8, so the values of z and y are dependent on each other.

Now, let's substitute the values of z and y back into Equation 2:

(z + w) = 16
(z + (8 - z)) = 16
8 + z = 16
z = 8

Now, we can substitute the value of z into Equation 4:

(y + w) = 8
(8 - z + w) = 8
(8 - 8 + w) = 8
w = 0

Now, let's substitute the values of z and w back into Equation 2:

(z + w) = 16
(8 + 0) = 16
8 = 16

As 8 does not equal 16, this is not possible, which means there is no solution to this system of equations that satisfies all the conditions.