Let \( x \) be the weight of a large box and \( y \) be the weight of a small box. We can set up the following system of equations based on the information provided:
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For the delivery of 3 large boxes and 5 small boxes: \[ 3x + 5y = 116 \quad \text{(Equation 1)} \]
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For the delivery of 9 large boxes and 7 small boxes: \[ 9x + 7y = 238 \quad \text{(Equation 2)} \]
We can solve this system of equations using the method of substitution or elimination. Here, we will use the elimination method.
First, we will multiply Equation 1 by 3 to match the coefficient of \( x \) in Equation 2:
\[ 3(3x + 5y) = 3(116) \] This simplifies to: \[ 9x + 15y = 348 \quad \text{(Equation 3)} \]
Now we have:
- Equation 2: \( 9x + 7y = 238 \)
- Equation 3: \( 9x + 15y = 348 \)
Next, we subtract Equation 2 from Equation 3 to eliminate \( x \):
\[ (9x + 15y) - (9x + 7y) = 348 - 238 \] This simplifies to: \[ 15y - 7y = 110 \] \[ 8y = 110 \] \[ y = \frac{110}{8} = 13.75 \]
Now that we have \( y \), we can substitute back into Equation 1 to find \( x \):
\[ 3x + 5(13.75) = 116 \] \[ 3x + 68.75 = 116 \] \[ 3x = 116 - 68.75 \] \[ 3x = 47.25 \] \[ x = \frac{47.25}{3} = 15.75 \]
Now we have the weights of the boxes:
- The weight of a large box \( x \) is 15.75 kilograms.
- The weight of a small box \( y \) is 13.75 kilograms.
Thus, the final answer is:
- Large box weight: 15.75 kg
- Small box weight: 13.75 kg
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