QUESTION IMAGE
Question
| first delivery | second delivery | |
|---|---|---|
| number of small boxes | 2 | 5 |
| total weight (in kilograms) | 199 | 79 |
let x be the weight (in kilograms) of each large box.
let y be the weight (in kilograms) of each small box.
(a) write a system of equations that could be used to find the weight (in kilograms) of each type of box.
□x + □y = □
□x + □y = □
(b) how much does each type of box weigh (in kilograms)?
weight of each large box: □ kilograms
weight of each small box: □ kilograms
Part (a)
Step1: Analyze first delivery
For the first delivery, there are 12 large boxes (each of weight \( x \)) and 2 small boxes (each of weight \( y \)), with a total weight of 199 kg. So the equation is \( 12x + 2y = 199 \).
Step2: Analyze second delivery
For the second delivery, there are 3 large boxes (each of weight \( x \)) and 5 small boxes (each of weight \( y \)), with a total weight of 79 kg. So the equation is \( 3x + 5y = 79 \).
Step1: Solve the system of equations
We have the system:
\[
\]
Multiply equation (2) by 4 to get: \( 12x + 20y = 316 \quad (3) \)
Subtract equation (1) from equation (3):
\( (12x + 20y) - (12x + 2y) = 316 - 199 \)
\( 18y = 117 \)
\( y = \frac{117}{18} = 6.5 \)
Step2: Substitute \( y \) into equation (2)
Substitute \( y = 6.5 \) into \( 3x + 5y = 79 \):
\( 3x + 5(6.5) = 79 \)
\( 3x + 32.5 = 79 \)
\( 3x = 79 - 32.5 = 46.5 \)
\( x = \frac{46.5}{3} = 15.5 \)
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\( 12x + 2y = 199 \)
\( 3x + 5y = 79 \)