Question

A restaurant manager buys 50 pounds of sausage and 80 pounds of hamburger for a total cost of $300. A second purchase, at the same price, includes 100 pounds of sausage and 120 pounds of hamburger for a cost of $480. Find the cost of sausage and the hamburger.

Answer 1

Sausage cost = $1.20; hamburger cost = $3.00

Explanation:

Determining the cost of each sausage and e

• We will need a system of equations to find the cost of each sausage and each hamburger.
• We can let s represent the cost of each sausage and let h represent the cost of each hamburger.

For both purchases, we know that the sum of the costs of the sausages and hamburgers equals the total cost:

(sausage amount * cost) + (hamburger amount * cost) = total cost

First equation:

Since the manager's first purchase of 50 lbs of sausage and 80 lbs of hamburger resulted in a total cost of $300, our first equation is given by:

50s + 80h = 300

Second equation:

Since the manager's second purchase of 100 lbs of sausage and 120 lbs of hamburger resulted in a total cost of $480, our second equation is given by:

100s + 120h = 480

Method to solve: Elimination:

• We can start by multiplying the first equation by -2.

• Doing so will allow us to eliminate s since -100s + 100h = 0:

-2(50s + 80h = 300)

-100s - 160h = -600

Solving for h (the cost of each hamburger):

Now we can add the two equations to eliminate s and solve for h:

100s + 120h = 480

+

-100s - 160h = -600

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(100s - 100s) + (120h - 160h) = (480 - 600)

(-40h = -120) / -40

h = 3

Thus, each hamburger costs $3.00

Solving for s (the cost of each sausage):

Now we can solve for s by plugging in 3 for h in the first equation (50s + 80h = 300):

50s + 80(3) =300

(50s + 240 = 300) - 240

(50s = 60) / 50

s = 1.2

This, each sausage costs $1.20