Question

can someone help me? how do you ask for help?
Candy selling at $6.00 per pound will be mixed with candy selling at $9.00 per pound. How many pounds of the more expensive candy are needed to produce a 15-pound mixture that sells for $7.00 per pound?

Answer 1

@amistre64 i need help do I have to do something to get help? im not sure how to use openstudy

Answer 2

you post and either wait or tag, when people have the time they will respond.

Answer 3

what are your thoughts on this problem?

Answer 4

i thought I had to write the equations as fractions to equal 15 but it doesnt work

Answer 5

well lets talk our way thru it

x pounds of candy1, added to y pounds of candy, have give us 15 lbs together

is x + y = 15 a good start?

Answer 6

now

x pounds at $6 a pound costs 6x
y pounds at $9 a pound costs 9y

when combined, they should cost $7

6x + 9y = 7 is also a consideration to this problem

Answer 7

I know its somewhere between 3 and 4 pounds of the $9 candy

Answer 8

since x and y are the same, it is a solution to the system of both equation

x+y = 15
6x + 9y = 7

now lets say x = 7-n, and y=8+n

x = 7-n
+y = 8+n
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x+y = 15 for all x and y so lets use that in the other one




6x + 9y = 7

6(7-n) + 9(8+n) = 7

now we can solve for n and thereby know x and y

Answer 9

sorry I didnt see your post

Answer 10

this is just basic a idea, a strategy (one of many), does it make sense tho?

Answer 11

give me a sec

Answer 12

3.56

Answer 13

im not getting a realistic solution, at least not from the wolf.

Answer 14

x+y = 15

6x + 9y = 7

either the problem has no solution, or ther eis something wrong with the setup here

Answer 15

7(15) we have 15 pounds

Answer 16

x+y = 15

6x + 9y = 7(15)

thats better

Answer 17

oh I got it but probably not the right way
10 pounds of the $6 candy and 5 pounds of the $9 candy
60+45= 105 divided by 15 so it would be 7
not the right way to do it

Answer 18

your way is the right way

Answer 19

there is no 'right' or wrong way .. there are multiple methods that can reach the same conclusion, as long as we have the correct information to play with :)

10 of x, and 5 of y is fine

Answer 20

:)

Answer 21

hat math i have dyscalculia so any math is really hard for me to compute and takes me longer

Answer 22

I hate :P

Answer 23

i just old and forgetful lol

Answer 24

lol thanxs