Let's denote the cost of one pencil as \( p \) and the cost of one rubber as \( r \).
From the given information, we can set up the following system of equations based on the costs:
- \( 6p + 3r = 1170 \) (Equation 1)
- \( 5p + 2r = 920 \) (Equation 2)
We can solve this system of equations step-by-step.
First, we can simplify Equation 1 by dividing everything by 3:
\[ 2p + r = 390 \quad \text{(Equation 3)} \]
Next, we can multiply Equation 3 by 2 to align the coefficient of \( r \):
\[ 4p + 2r = 780 \quad \text{(Equation 4)} \]
Now, we will subtract Equation 2 from Equation 4:
\[ (4p + 2r) - (5p + 2r) = 780 - 920 \]
This simplifies to:
\[ -1p = -140 \]
So, we find:
\[ p = 140 \]
Now that we have \( p \), we can substitute it back into Equation 3 to find \( r \):
\[ 2(140) + r = 390 \]
This simplifies to:
\[ 280 + r = 390 \]
Then, solving for \( r \):
\[ r = 390 - 280 = 110 \]
Thus, the costs are:
- The cost of one pencil \( p \) is #140.
- The cost of one rubber \( r \) is #110.
Final answer: Pencil = #140, Rubber = #110.
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