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Question 1103041: 1 pound of cereal A contains 10g of potassium, 2g of calcium, and 4g of magnesium. 1 pound of cereal B contains 6g of potassium, 0g of calcium, and 3g of magnesium. 1 pound of cereal C contains 12g of potassium, 6g of calcium, and 2g of magnesium. How many pounds of each cereal is needed to have a mixture containing 98g of potassium, 28g of calcium, and 32g of magnesium?
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
Let x be the number of grams of cereal A, y the grams of cereal B, and z the grams of cereal C. Then
the total amount of potassium is 10x+6y+12z;
the total amount of calcium is 2x+6z; and
the total amount of magnesium is 4x+3y+2z.
The mixture needs to have 98g of potassium, 28g of calcium, and 32g of magnesium, so
(1) 
(2) 
(3) 
There are many standard ways to solve systems of equations like this. However, it is always worthwhile to look at the system of equations to see if there is a quick path to the solution using basic algebra.
The first two equations both have a common factor of 2, so we can divide them out, leaving
(1) 
(2) 
(3)
Now a quick look at these equations shows that adding (2) and (3) will give us an equation almost identical to (1), allowing us to immediately solve for one variable:
(1)
(4) 
giving us z = 3.
Then equation (2) gives us x = 5; and then either equation (1) or (3) gives us y = 2.
ANSWER: 5g of cereal A, 2g of cereal B, and 3g of cereal C
CHECK:
potassium: 10(5)+6(2)+12(3) = 50+12+36 = 98
calcium: 2(5)+6(3) = 10+18 = 28
magnesium: 4(5)+3(2)+2(3) = 20+6+6 = 32
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