SOLUTION: I need help with this word problem. Thanks in advance!
A goldsmith has two alloys that are different purities of gold. The first is three-fourths pure gold and the second is f
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A goldsmith has two alloys that are different purities of gold. The first is three-fourths pure gold and the second is f
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Question 128861This question is from textbook Introductory Algebra
: I need help with this word problem. Thanks in advance!
A goldsmith has two alloys that are different purities of gold. The first is three-fourths pure gold and the second is five-twelths pure gold. How many ounces of each should be melted and mixed in order to obtain a 6-oz. mixture that is two-thirds pure gold?
I have come up with:
First + Second = Mixture
x + y = 6 oz
3/4 + 5/12 = 2/3
3/4x + 5/12y = ?4 This question is from textbook Introductory Algebra
You can put this solution on YOUR website! A goldsmith has two alloys that are different purities of gold. The first is three-fourths pure gold and the second is five-twelths pure gold. How many ounces of each should be melted and mixed in order to obtain a 6-oz. mixture that is two-thirds pure gold?
I have come up with:
First + Second = Mixture
Let "x" be amount of (3/4) gold ; Let "y" be amount of (5/12) gold.
EQUATIONS:
Weight Equation: x + y = 6 oz
Gold Equation: (3/4)x + (5/12)y = (2/3)6 = 4 oz
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Rearrange:
x = 6-y
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Substitute to solve for y:
(3/4)(6-y) + (5/12)y = 4
Multiply thru by 12 to get:
9(6-y) + 5y = 48
54-9y + 5y = 48
-4y = -6
y = 3/2 (amount of 5/12 gold)
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Substitute into x = 6-y to solve for "x":
x = 6-(3/2)= 9/2 (amount of 3/4 gold)
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Cheers,
Stan H.
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