Edwin works at a fast-food restaurant. Every Tuesday and Friday, he helps unload a delivery truck. This table shows how many boxes of hamburger buns and how many boxes of potatoes he unloaded last week.

To find the weight of each box of hamburger buns and potatoes, we can use a system of equations based on the information provided.

Let's define:

• \( b \) = weight of each box of buns (in pounds)
• \( p \) = weight of each box of potatoes (in pounds)

From the table:

1. On Tuesday, Edwin unloaded 40 boxes of buns and 35 boxes of potatoes with a total weight of 1,945 pounds: \[ 40b + 35p = 1945 \quad (1) \]

2. On Friday, he unloaded 60 boxes of buns and 70 boxes of potatoes with a total weight of 3,530 pounds: \[ 60b + 70p = 3530 \quad (2) \]

Now we have the following two equations to solve:

\[ 40b + 35p = 1945 \tag{1} \] \[ 60b + 70p = 3530 \tag{2} \]

We can simplify equation (2) by dividing everything by 10:

\[ 6b + 7p = 353 \tag{3} \]

Now, we will solve equations (1) and (3) together. First, let's multiply equation (1) by 6 and equation (3) by 4 to align the coefficients of \( b \):

\[ 240b + 210p = 11670 \tag{4} \quad (from , equation , 1) \] \[ 24b + 28p = 1412 \tag{5} \quad (from , equation , 3) \]

Next, we can multiply equation (5) by 10 to further align coefficients for subtraction:

\[ 240b + 280p = 14120 \tag{6} \]

Now we will subtract equation (4) from equation (6):

\[ (240b + 280p) - (240b + 210p) = 14120 - 11670 \]

This simplifies to:

\[ 70p = 2450 \]

Solving for \( p \):

\[ p = \frac{2450}{70} = 35 \]

Now we have \( p = 35 \) pounds for each box of potatoes. We can substitute \( p \) back into equation (1) to find \( b \):

\[ 40b + 35(35) = 1945 \] \[ 40b + 1225 = 1945 \] \[ 40b = 1945 - 1225 \] \[ 40b = 720 \] \[ b = \frac{720}{40} = 18 \]

So, the weights of the boxes are:

• A box of buns weighs 18 pounds.
• A box of potatoes weighs 35 pounds.