Let's define the variables for both dealerships based on the problem statement:
- Let \( c_A \) be the number of cars sold by Dealership A.
- Let \( t_A \) be the number of trucks sold by Dealership A.
- Let \( c_B \) be the number of cars sold by Dealership B.
- Let \( t_B \) be the number of trucks sold by Dealership B.
a) Equation for the total cars and trucks for Dealership A:
From the problem, we know:
- Dealership A sells a total of 164 cars and trucks.
Thus, the equation for Dealership A is: \[ c_A + t_A = 164 \]
b) Equation for the total cars and trucks for Dealership B:
From the problem, we learn that Dealership B sells twice as many cars and half as many trucks as Dealership A and has a total of 229 cars and trucks.
Thus, we have:
- \( c_B = 2c_A \)
- \( t_B = \frac{1}{2}t_A \)
The equation for Dealership B is: \[ c_B + t_B = 229 \] Substituting \( c_B \) and \( t_B \): \[ 2c_A + \frac{1}{2}t_A = 229 \]
c) How many cars did Dealership A sell?
Now we can solve the two equations simultaneously.
We have:
- \( c_A + t_A = 164 \) (Equation 1)
- \( 2c_A + \frac{1}{2}t_A = 229 \) (Equation 2)
From Equation 1, we can express \( t_A \) in terms of \( c_A \): \[ t_A = 164 - c_A \]
Now substitute \( t_A \) into Equation 2: \[ 2c_A + \frac{1}{2}(164 - c_A) = 229 \] Distributing \(\frac{1}{2}\): \[ 2c_A + 82 - \frac{1}{2}c_A = 229 \]
To eliminate the fraction, multiply each term by 2: \[ 4c_A + 164 - c_A = 458 \] Combine like terms: \[ 3c_A + 164 = 458 \] Subtract 164 from both sides: \[ 3c_A = 294 \] Now, divide by 3: \[ c_A = 98 \]
d) How many trucks did Dealership B sell?
Using the value of \( c_A \) to find \( t_A \): \[ t_A = 164 - c_A = 164 - 98 = 66 \]
Now, to find the number of trucks sold by Dealership B: \[ t_B = \frac{1}{2} t_A = \frac{1}{2} \cdot 66 = 33 \]
Summary of Answers:
a) \( c_A + t_A = 164 \)
b) \( 2c_A + \frac{1}{2}t_A = 229 \)
c) Dealership A sold 98 cars.
d) Dealership B sold 33 trucks.
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