Let's denote the cost of one pencil as \( p \) and the cost of one rubber as \( r \).
From the information given, we can set up the following equations based on the costs:
- \( 6p + 3r = 1170 \) (Equation 1)
- \( 5p + 2r = 920 \) (Equation 2)
Now, we can solve these equations simultaneously.
First, let's simplify Equation 1. We can divide the entire equation by 3:
\[ 2p + r = 390 \quad \text{(Equation 3)} \]
Next, we can use Equation 3 to express \( r \) in terms of \( p \):
\[ r = 390 - 2p \]
Now we'll substitute this expression for \( r \) back into Equation 2:
\[ 5p + 2(390 - 2p) = 920 \]
Expanding this gives:
\[ 5p + 780 - 4p = 920 \]
Now, combine like terms:
\[ p + 780 = 920 \]
Subtract 780 from both sides:
\[ p = 140 \]
Now we know the cost of one pencil is \( p = 140 \). Next, we'll substitute \( p \) back into Equation 3 to find \( r \):
\[ 2(140) + r = 390 \]
This simplifies to:
\[ 280 + r = 390 \]
Subtract 280 from both sides:
\[ r = 110 \]
Hence, the costs are:
- Cost of one pencil: \( #140 \)
- Cost of one rubber: \( #110 \)
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