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Question
QUESTION: Two separate piles contain mixtures of sand and gravel.

Pile A- Mixture is 1:3 ratio

Pile B- Mixture is 3:5 ratio

Pile C is to contain 5 cubic yards of sand and 9 cubic yards of gravel.

Calculate amount used from Pile A and B.  

Thanks.

ANSWER: "Pile A- Mixture is 1:3 ratio of sand to gravel."
1 + 3 = 4
1 yd³ from Pile A contains ¼ yd³ of sand and ¾ yd³ of gravel.

"Pile B- Mixture is 3:5 ratio of sand to gravel."
3 + 5 = 8
1 yd³ from Pile B contains ⅜ yd³ of sand and ⅝ yd³ of gravel

"Pile C is to contain 5 cubic yards of sand and 9 cubic yards of gravel."
Let a be the cubic yards used from Pile A , and b the cubic yards used from Pile B.
(¼)a + (⅜)b = 5
(¾)a + (⅝)b = 9

You have a system of two equations with two unknowns. For now, I'll assume that you can solve it.


---------- FOLLOW-UP ----------

QUESTION: "(¼)a + (⅜)b = 5
(¾)a + (⅝)b = 9"

(1/4)a * 2/2 + (3/8)b = 5

(2/8)a + (3/8)b =  10


(3/4)a * 2/2 + (3/8)b = 9

(6/8)a + (3/8)b = 18

I am not sure.  Thanks again.

Answer
To solve for two unknowns, you need two equations, which is what you have. However, I just thought of a third equation. Since Pile C will contain a total of 14 yd³:
 a+b = 14
or
 a = 14-b

Now you can substitute 14-b for a in one of the other two equations, then solve for b. (This may be easier to read: https://www.flickr.com/photos/dwread/14494488085/ )
(1/4)(14-b) + (3/8)b = 5
(7/2) - (1/4)b + (3/8)b = 5
-(1/4)b + (3/8)b = 5 - 7/2
(1/8)b = 3/2
b = 12
a = 14 - b = 2


Pile C contains 2 yd³ from Pile A and 12 yd³ from Pile B.

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