A landscaping company placed two orders with a nursery. The first order was for 13 bushes and 4 trees, and totaled $487.00. The second was for 6 bushes and 2 trees, and totaled $232.00. The bills do not list the per-item price. What were the costs of one bush and of one tree?
Looks like algebra I to me. Anyway,
(1) 13b+4t = 487
(2) 6b+2t = 232
Subtracting twice #2 from #1 we get
b = 23
so, t = 47
b+t = 70
To find the costs of one bush and one tree, we can use a system of two equations. Let's assign variables to the costs of one bush and one tree. Let's say the cost of one bush is B dollars, and the cost of one tree is T dollars.
From the information given, we can create the following equations:
Equation 1: 13B + 4T = 487.00 (total cost of the first order)
Equation 2: 6B + 2T = 232.00 (total cost of the second order)
We now have a system of two equations with two variables. We can solve this system using substitution or elimination methods. Let's use the elimination method:
Multiply Equation 1 by 2 and Equation 2 by -4 to eliminate the T variable:
(2) (13B + 4T) = (2) (487.00) becomes 26B + 8T = 974.00
(-4) (6B + 2T) = (-4) (232.00) becomes -24B - 8T = -928.00
Now, add the two equations:
26B + 8T + (-24B) - 8T = 974.00 + (-928.00)
2B = 46.00
Divide both sides of the equation by 2 to solve for B:
2B / 2 = 46.00 / 2
B = 23.00
Now that we know the cost of one bush (B = 23.00), we can substitute this value into Equation 1 to find the cost of one tree:
13B + 4T = 487.00
13(23.00) + 4T = 487.00
299.00 + 4T = 487.00
Subtract 299.00 from both sides of the equation:
4T = 488.00 - 299.00
4T = 188.00
Divide both sides of the equation by 4 to solve for T:
4T / 4 = 188.00 / 4
T = 47.00
Therefore, the cost of one bush is $23.00, and the cost of one tree is $47.00.
c=15+10h
c=8h