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A student government organization is selling Christmas trees as a fundraiser. On Friday, they sold 5 noble fir trees and 3 douglas fir trees for a total of $420. On Saturday, 12 noble fir trees and 9 douglas fir trees were sold for a total of $1,080. What is the cost per tree for each type?



Answer :

collinse
So you must set this up as a system of equations
n= noble fir trees
d= douglas fir trees
5n + 3d = $420
12n + 9d = $1,080 
To get a positive output you have to multiply the second equation by 3 to be able to be divisable to one another 
36n + 18d = 3,240
now divide the equation by 18 
2n + d = 180
-2n        -2n
d = 180 - 2n
Now put that equation in the first as d ^^^
5n + 3(180 -2n) = 420
5n + 540- 6n = 420 
      - 540         -540 
1n = 120 
n = 120 
Now put n into the second equation( none changed one) to find d 
12(120) + 9d = 3240 
1440 + 9d = 3240 
-1440        -1440
9d = 1800
÷9    ÷9
d= 200
so the noble is $ 120 
and the douglas is $200

Answer:

The cost of noble fir tree is $60.

The cost of douglas fir tree is $40.

Step-by-step explanation:

Given : A student government organization is selling Christmas trees as a fundraiser.

To find : What is the cost per tree for each type?

Solution :

Let x be the cost of noble fir tree.

and y be the cost of douglas fir tree.

On Friday, they sold 5 noble fir trees and 3 douglas fir trees for a total of $420.

i.e. [tex]5x + 3y = 420[/tex] ....(1)

On Saturday, 12 noble fir trees and 9 douglas fir trees were sold for a total of $1,080.

i.e. [tex]12x + 9y = 1080[/tex]

[tex]4x +3y =360[/tex] ......(2)

Solving equation (1) and (2),

Subtract (2) from (1),

[tex]5x + 3y-(4x +3y)= 420-360[/tex]

[tex]5x + 3y-4x-3y=60[/tex]

[tex]x=60[/tex]

Substitute in equation (1),

[tex]5(60)+ 3y = 420[/tex]

[tex]300+ 3y = 420[/tex]

[tex]3y=120[/tex]

[tex]y=40[/tex]

The cost of noble fir tree is $60.

The cost of douglas fir tree is $40.

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