To solve this problem, we need to determine the value of x, which represents the number of plates per setting.
Let's break down the information given:
Weight of each piece of silverware: 4 ounces
Number of pieces of silverware per setting: 3
Weight of each plate: 12 ounces
Number of plates per setting: x
Total weight of all the settings: 5040 ounces
Number of tables: 15
Number of settings per table: 8
Number of backup settings: 20
We can set up the equation as follows:
(Number of tables * Number of settings per table * (Weight of each piece of silverware * Number of pieces of silverware per setting + Weight of each plate * Number of plates per setting)) + (Number of backup settings * (Weight of each piece of silverware * Number of pieces of silverware per setting + Weight of each plate * Number of plates per setting)) = Total weight of all the settings
(15 * 8 * (4 * 3 + 12 * x)) + (20 * (4 * 3 + 12 * x)) = 5040
Let's simplify and solve for x:
(15 * 8 * (12 + 12x)) + (20 * (12 + 12x)) = 5040
(15 * 8 * 12 + 15 * 8 * 12x) + (20 * 12 + 20 * 12x) = 5040
(1440 + 1440x) + (240 + 240x) = 5040
To isolate x, we combine like terms:
1440 + 1440x + 240 + 240x = 5040
1680 + 1680x = 5040
To isolate x, we subtract 1680 from both sides of the equation:
1680 + 1680x - 1680 = 5040 - 1680
1680x = 3360
To solve for x, we divide both sides of the equation by 1680:
x = 3360 / 1680
x = 2
The value of x is 2.
#### 2
The answer is: 2
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Mason is a caterer packing up silverware and plates for a big corporate event. Each piece of silverware weighs 4 ounces, and there are three pieces of silverware per setting. Each plate weighs 12 ounces, and there are x plates per setting. If Mason needs enough settings for 15 tables with 8 settings each, plus 20 backup settings in case of breakage, All the settings will weigh 5040 ounces.

### Answer:
