Jerrica is packing a box that can hold at most 65 pounds. She starts by putting a container in the box that weighs 26 pounds. She plans to fill the rest of the box with books. Each book she places in the box weighs 1.3 pounds. Write an inequality for the number of books Jerrica can put in the box and describe the solutions. (2 points) 1.3x+26≤ 65,

Question

where x is the number of books Jerrica can put in the box. Jerrica can put at most f books in the box.

Answer 1

To write the inequality for the number of books Jerrica can put in the box, we start with the total weight that the box can hold, which is 65 pounds.

1. Jerrica puts a container that weighs 26 pounds in the box.
2. Each book weighs 1.3 pounds.
3. Let \( x \) be the number of books Jerrica can put in the box.

The total weight of the container and the books combined should not exceed the box's maximum capacity. Therefore, the inequality can be set up as follows:

\[ 1.3x + 26 \leq 65 \]

Now, we can solve this inequality to find the maximum number of books \( x \) that Jerrica can put in the box:

1. Subtract 26 from both sides:

\[ 1.3x \leq 65 - 26 \] \[ 1.3x \leq 39 \]

2. Divide both sides by 1.3:

\[ x \leq \frac{39}{1.3} \]

3. Calculate \( \frac{39}{1.3} \):

\[ x \leq 30 \]

Thus, the inequality \( 1.3x + 26 \leq 65 \) constrains \( x \) to be \( x \leq 30 \).

Description of the solutions:

• The solutions to the inequality represent the possible values of \( x \), which is the maximum number of books Jerrica can place in the box.
• Since \( x \) must be a non-negative integer (as she can't put a negative number of books), the feasible integer values for \( x \) are 0 through 30.
• Therefore, Jerrica can put at most 30 books in the box.