SOLUTION: "To Solve Value Mixture Problems" A goldsmith combined an alloy that cost $4.30 per ounce with an alloy that cost $1.80 per ounce. How many ounces of each were used to make a mi

Question 280725: "To Solve Value Mixture Problems"
A goldsmith combined an alloy that cost $4.30 per ounce with an alloy that cost $1.80 per ounce. How many ounces of each were used to make a mixture of 200 oz costing $2.50 per ounce ?
Thanks
Answer by Grinnell(63) (Show Source): You can put this solution on YOUR website!
ok.
Let us set it up.
The total number of ounces will be 200.
There will be a number of ounces at $4.30.
And, a number of ounces at $1.80.
Let x represent the number of ounces at $1.80.
Let 200-x represent the number of ounces at $4.30.
Off to the races...
(200-x)$4.30+$1.80x=200 TIMES $2.50
First let's multiply by 100 to get rid of the decimal.
(200-x)430+180x=200 TIMES 250
(Review the distributive property of multiplication)
We get 86,000-430x+180x=50,000
-430x+180x=-36,000
-250x=-36,000
x=144, go back and plug in! (What did x represent?) 144ounces at $1.80/ therefore 56ounces at $4.30.
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