Let \( c \) be the cost to rent one chair and \( t \) be the cost to rent one table. We can set up the following system of equations based on the information given:
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From the first scenario:
\( 7c + 9t = 91 \) \quad (Equation 1) -
From the second scenario:
\( 5c + 3t = 35 \) \quad (Equation 2)
Now we can solve this system of equations.
First, let's solve Equation 2 for one of the variables. We'll solve for \( c \):
\[ 5c + 3t = 35 \]
Subtract \( 3t \) from both sides:
\[ 5c = 35 - 3t \]
Now divide by 5:
\[ c = \frac{35 - 3t}{5} \quad (Equation 3) \]
Next, substitute Equation 3 into Equation 1:
\[ 7\left(\frac{35 - 3t}{5}\right) + 9t = 91 \]
Multiply everything by 5 to eliminate the fraction:
\[ 7(35 - 3t) + 45t = 455 \]
Distribute the 7:
\[ 245 - 21t + 45t = 455 \]
Combine like terms:
\[ 245 + 24t = 455 \]
Now, subtract 245 from both sides:
\[ 24t = 210 \]
Now divide by 24:
\[ t = \frac{210}{24} = \frac{35}{4} = 8.75 \]
Now that we have the cost of a table, substitute \( t \) back into Equation 3 to find \( c \):
\[ c = \frac{35 - 3(8.75)}{5} \]
Calculate \( 3(8.75) \):
\[ 3(8.75) = 26.25 \]
Now substitute it back into the equation for \( c \):
\[ c = \frac{35 - 26.25}{5} = \frac{8.75}{5} = 1.75 \]
So, the costs are:
\[ \text{Cost of each chair } c = 1.75 \] \[ \text{Cost of each table } t = 8.75 \]
Thus, the cost to rent each chair is \( \boxed{1.75} \) and the cost to rent each table is \( \boxed{8.75} \).
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