SOLUTION: The question is "How many ounces of a silver alloy that costs $11 per ounce must be mixed with 9 ounces of a silver alloy that costs $8 per ounce to make an alloy that costs $9 per

Algebra ->  Equations -> SOLUTION: The question is "How many ounces of a silver alloy that costs $11 per ounce must be mixed with 9 ounces of a silver alloy that costs $8 per ounce to make an alloy that costs $9 per      Log On


   

Question 942578: The question is "How many ounces of a silver alloy that costs $11 per ounce must be mixed with 9 ounces of a silver alloy that costs $8 per ounce to make an alloy that costs $9 per ounce?"
The problem I have is understanding what formula to use and exactly how to you solve it. I would appreciate the help thank you!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +x+ = ounces of $11/ounce alloy to be used
-------------
+11x+%2B+8%2A9+=+9%2A%28+x+%2B+9+%29+
+11x+%2B+72+=+9x+%2B+81+
+2x+=+9+
+x+=+4.5+
4.5 ounces of $11/ounce alloy should be used
-------------
check:
+11x+%2B+8%2A9+=+9%2A%28+x+%2B+9+%29+
+11%2A4.5+%2B+8%2A9+=+9%2A%28+4.5+%2B+9+%29+
+49.5+%2B+72+=+9%2A13.5+
+121.5++=+121.5+
OK