SOLUTION: two metal alloys, containing 26% copper and 54% copper are to be mixed together to get 210 oz that is 30% copper. How much of each should be used?

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Question 644755: two metal alloys, containing 26% copper and 54% copper are to be mixed together to get 210 oz that is 30% copper. How much of each should be used?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = ounces of 1st alloy needed
Let +b+ = ounces of 2nd alloy needed
given:
+.26a+ = ounces of copper in 1st alloy
+.54b+ = ounces of copper in 2nd alloy
(1) +%28+.26a+%2B+.54b+%29+%2F+210+=+.3+
(2) +a+%2B+b+=+210+
-----------------
(1) +.26a+%2B+.54b+=+.3%2A210+
(1) +.26a+%2B+.54b+=+63+
(1) +26a+%2B+54b+=+6300+
Multiply both sides of (2) by +26+
and subtract (2) from (1)
(1) +26a+%2B+54b+=+6300+
(2) +-26a+-+26b+=+-5460+
+28b+=+840+
+b+=+30+
and, since
(2) +a+%2B+b+=+210+
(2) +a+=+210+-+30+
(2) +a+=+180+
180 ounces of 1st alloy are needed
30 ounces of 2nd alloy are needed
check:
(1) +%28+.26a+%2B+.54b+%29+%2F+210+=+.3+
(1) +%28+.26%2A180+%2B+.54%2A30+%29+%2F+210+=+.3+
(1) +46.8+%2B+16.2+=+.3%2A210+
(1) +63+=+63+
OK