SOLUTION: six apples and three oranges cost $3.36.two apples and five oranges cost $3.04.find the cost of an apple and the cost of an orange.

Algebra ->  Systems-of-equations -> SOLUTION: six apples and three oranges cost $3.36.two apples and five oranges cost $3.04.find the cost of an apple and the cost of an orange.      Log On


   



Question 118929: six apples and three oranges cost $3.36.two apples and five oranges cost $3.04.find the cost of an apple and the cost of an orange.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Let apples be x and oranges be y
If six apples and three oranges cost $3.36, we have:
6x+%2B+3y+=+3.36
If two+ apples and +five+ oranges cost $3.04, we have:
2x+%2B+5y+=+3.04

Now solve the system:

6x+%2B+3y+=+3.36…………………(1)….solve for x and substitute in (2)
2x+%2B+5y+=+3.04…………………(2)

6x++=++-+3y+%2B+3.36………………...divide both sides by 6
6x%2F6++=++-+3y%2F6+%2B+3.36%2F6….

x++=++-+%281%2F2%29y+%2B+0.56……………. substitute in (2)


2%28-+%281%2F2%29y+%2B+0.56%29+%2B+5y+=+3.04

cross%282%29%28-+%281%2Fcross%282%29%29%29y+%2B+2%2A0.56+%2B+5y+=+3.04

-y+%2B+1.12%2B+5y+=+3.04

+1.12%2B+4y+=+3.04

+4y+=+3.04+-1.12

+4y+=+1.92……………………….. divide both sides by 4

+4y%2F4+=+1.92%2F4………………………..

+y+=+0.48………………………..now find x

x++=++-+%281%2F2%29y+%2B+0.56…………….

x++=++-+%281%2F2%29+%280.48%29+%2B+0.56…………….

x++=++%28-0.24%29+%2B+0.56…………….

x++=++0.32…………….
Check:
6x+%2B+3y+=+3.36

6%280.32%29+%2B+3%280.48%29+=+3.36

1.92%2B+1.44+=+3.36

3.36+=+3.36