Gerald owns a livestock trailer that can hold up to 10,000 pounds of livestock. He needs to transport 36 goats and 18 calves in his livestock trailer in one trip. Assume the trailer has enough room to transport this many livestock.

Question

Let g represent the average weight of each goat in pounds and let c represent the average weight of each calf in pounds.

Create the inequality that symbolizes this situation, and then answer the question below.

blank g + blank c ≤

Can Gerald transport these goats and calves in his livestock trailer in one trip if the average weight of each goat is 96 pounds and the average weight of each calf is 369 pounds?

Answer 1

To create the inequality that represents this situation, we will consider the total weight of the goats and calves that Gerald wants to transport.

Let:

• \( g \) represent the average weight of each goat (in pounds).
• \( c \) represent the average weight of each calf (in pounds).

Gerald has:

• 36 goats, each weighing \( g \) pounds, which gives a total weight of \( 36g \) pounds.
• 18 calves, each weighing \( c \) pounds, which gives a total weight of \( 18c \) pounds.

The total weight of the livestock that Gerald wants to transport should not exceed the trailer's weight limit of 10,000 pounds. Thus, the inequality can be set up as follows:

\[ 36g + 18c \leq 10,000 \]

Now let's substitute \( g = 96 \) pounds and \( c = 369 \) pounds into the inequality:

\[ 36(96) + 18(369) \leq 10,000 \]

Calculating the left side:

1. Calculate \( 36(96) \): \[ 36 \times 96 = 3456 \]

2. Calculate \( 18(369) \): \[ 18 \times 369 = 6642 \]

Now add these two results together: \[ 3456 + 6642 = 10098 \]

Now we check the inequality: \[ 10098 \leq 10,000 \]

Since \( 10098 \) is greater than \( 10,000 \), the inequality is not satisfied.

Therefore, Gerald cannot transport these goats and calves in his livestock trailer in one trip.