A car manufacturing company (the Company) can produce a car for $3000.00 and a truck for $5000.00. Suppose "C" cars and "T" trucks are sold to the dealer at mark-ups, respectively, of 20 and 30 percent. If the Company made a profit of $27 million on sales of $137 million in one particular year, how many cars and trucks were sold by the company in that year?

I need to make two equations out of this and then use the substitution or elimination method. I posted this question earlier but the two equations didn't work out because I kept getting a negative answer

5000(1.2)T+3000(1.3)C=137,000,000

5000*.2T+ 3000(.3)C=27,000,000

Which reduces to
6T+3.9C=137000
T+.9C=27000

Check my thinking.

Let's assume that "C" represents the number of cars sold and "T" represents the number of trucks sold by the company in that year.

The cost of producing a car is $3000, and the dealer markup is 20%. Therefore, the selling price of each car would be:
Selling price of a car = $3000 + (20% of $3000)
= $3000 + (0.20 * $3000)
= $3000 + $600
= $3600

Similarly, the cost of producing a truck is $5000, and the dealer markup is 30%. Therefore, the selling price of each truck would be:
Selling price of a truck = $5000 + (30% of $5000)
= $5000 + (0.30 * $5000)
= $5000 + $1500
= $6500

Now, the company made a profit of $27 million on sales of $137 million. We can express this information using the following equation:
(Cars sold * selling price of a car) + (Trucks sold * selling price of a truck) = Total sales - Company profit
(C * $3600) + (T * $6500) = $137,000,000 - $27,000,000
3600C + 6500T = 137,000,000 - 27,000,000
3600C + 6500T = 110,000,000 ----------------------------------(Equation 1)

We also know that the company sold a total of "C" cars and "T" trucks. Therefore, the total number of vehicles sold can be expressed as:
Total vehicles sold = Cars sold + Trucks sold
C + T = Total vehicles sold ----------------------------------------(Equation 2)

Now we have two equations:
3600C + 6500T = 110,000,000 ----------------------------------(Equation 1)
C + T = Total vehicles sold ----------------------------------------(Equation 2)

We can solve these equations using either the substitution or elimination method to find the values of "C" and "T."

Let's start by setting up the equations based on the given information.

The first equation represents the cost of producing and selling the vehicles, including the mark-up:

Cost = (Number of cars * (Cost per car + Mark-up on car)) + (Number of trucks * (Cost per truck + Mark-up on truck))

The second equation represents the total sales and profit:

Sales = (Number of cars * Selling price per car) + (Number of trucks * Selling price per truck)
Profit = Sales - Cost

Using the given values and information:

Cost per car = $3000
Mark-up on car = 20% = 0.20
Cost per truck = $5000
Mark-up on truck = 30% = 0.30
Profit = $27 million = $27,000,000
Sales = $137 million = $137,000,000

First, let's substitute the values into the equations:

1) Cost = (C * (3000 + 0.20*3000)) + (T * (5000 + 0.30*5000))
2) Sales = (C * Selling price per car) + (T * Selling price per truck)
Profit = Sales - Cost

Now, let's calculate the selling prices per vehicle:

Selling price per car = Cost per car + Mark-up on car = 3000 + 0.20*3000 = 3600
Selling price per truck = Cost per truck + Mark-up on truck = 5000 + 0.30*5000 = 6500

Substitute these values into the equations:

1) Cost = (C * 3600) + (T * 6500)
2) Sales = (C * 3600) + (T * 6500)
Profit = Sales - Cost = 27000000

Now, you have two equations:

1) (C * 3600) + (T * 6500) = Cost
2) (C * 3600) + (T * 6500) = Sales
Profit = 27000000

You can solve these equations using either the substitution or elimination method. I will demonstrate the substitution method.

Take equation 1 and solve for Cost:

Cost = (C * 3600) + (T * 6500)

Now, substitute this into equation 2:

(C * 3600) + (T * 6500) = Sales

Substitute Cost = (C * 3600) + (T * 6500) into equation 2:

(C * 3600) + (T * 6500) = 137000000

Simplify the equation:

3600C + 6500T = 137000000

Now, you have one equation in terms of C and T. Combine this with the profit equation:

3600C + 6500T = 137000000
Profit = 27000000

You can solve these two equations simultaneously using the substitution or elimination method to find the values of C and T, representing the number of cars and trucks sold by the company in that year.