mcruz65 said:
Code:
A goldsmith has two alloys that are different purities of gold. The first is three-fourths pure gold and the second is five-twelfths 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?
x + y = 6
3/4 + 5/12 = 2/3 or .75 + .42 = .68[color=#FF0000]<--------WHAT??? I certainly don' t think that .75 + .42 = .68!!!![/color]
x + y = 6 (Multiply both sides of this equation by -.75)
.75x + .42y = .68
-.75x - .75y = -4.5
.75x + .42y = .68
-------------- -----
-.33y -3.82
-.33y -3.82
---- = -----
-.33 -.33
y = 11.58
.75x + .42(11.58) = .68
.75x + 4.86 = .68
.75x + 4.86 - 4.86 = .68 -4.86
.75x = -4.18
.75x/.75 = -4.18/.75
x = -5.57
Insert the x and y values into the equation to solve the problem.
.75(-5.57) + .42(11.58) = .68
-4.18 + 4.86 = .68
x + y = 6
-5.57 + 11.58 = 6
Is this correct???
You did not define your variables.
I would do it this way:
let x = number of ounces of first alloy
let y = number of ounces of second alloy
You want to end up with 6 ounces....so,
x + y = 6 (you had this equation correct)
Now, the first alloy is 3/4 gold, so if you have x ounces of this alloy, you've got (3/5)x ounces of gold
The second alloy is 5/12 gold, so if you have y ounces of this alloy, you've got (5/12)y ounces of gold.
The mixture, of which you have 6 ounces, is to be 2/3 gold. So the amount of gold in the mixture is (2/3)*6
gold in the first alloy + gold in the second alloy = gold in the final mixture
(3/4)x + (5/12)y = (2/3)*6
or,
(3/4)x + (5/12)y = 4
Now...you have two equations in two variables....
x + y = 6
(3/4)x + (5/12)y = 4
Can you solve this system? AND....I surely would NOT use decimals!!!!
Here's a hint to make solving the system easier. Multiply both sides of the second equation by the common denominator of the fractions, which is 12:
12*(3/4)x + 12*(5/12)y = 12*4
9x + 5y = 48
Your two equations now are
x + y = 6
9x + 5y = 48
