Luke was buying toy cars and trucks. If he bought 2000 toy cars and 2012 toy trucks, he would need $9036. If he bought 2006 toy cars and 2006 toy trucks, he would need $9027. How much would 1 toy truck costs?
Let the cost of one toy car be x and the cost of one toy truck be y.
From the information given, we can set up the following equations:
2000x + 2012y = 9036 ...(1)
2006x + 2006y = 9027 ...(2)
To solve these equations, we will use the method of elimination or substitution.
First, let's eliminate x by multiplying equation (2) by (-1):
-2006x - 2006y = -9027
Now we can add this equation to equation (1) to eliminate x:
(2000x + 2012y) + (-2006x - 2006y) = 9036 + (-9027)
6y = 9
Divide both sides of the equation by 6 to solve for y:
y = 9/6 = 3/2 = $1.50
Therefore, 1 toy truck costs $1.50.
Let's assume the cost of one toy car is represented by "C" and the cost of one toy truck is represented by "T".
From the given information, we can create two equations:
Equation 1: 2000C + 2012T = 9036
Equation 2: 2006C + 2006T = 9027
To find the cost of one toy truck, we need to solve these equations simultaneously. We can use the method of substitution or elimination to find the value of "T".
Let's use the substitution method:
From Equation 1:
2000C + 2012T = 9036
=> 2012T = 9036 - 2000C
=> T = (9036 - 2000C)/2012
Substituting this value of "T" into Equation 2:
2006C + 2006((9036 - 2000C)/2012) = 9027
Simplifying this equation:
2006C + (9036 - 2000C) = 9027 * (2012/2006)
=> 2006C + 9036 - 2000C = 9037.828
Combining like terms:
6C + 9036 - 2000C = 9037.828
=> -1994C = 9037.828 - 9036
=> -1994C = 1.828
Dividing both sides by -1994:
C = 1.828 / -1994
=> C = -0.000916749
Since the value of "C" is negative, it means there is an error in the equations or the given information. Please double-check the values and make sure they are accurate.
However, if we assume that the value of "C" should be positive, we can conclude that there is an inconsistency in the given information.