SOLUTION: A mixture containing 11 % salt is to be mixed with 4 ounces of a mixture that is 20 % salt, in order to obtain a solution that is 17 % salt. How much of the first solution must be

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Question 902646: A mixture containing 11 % salt is to be mixed with 4 ounces of a mixture that is 20 % salt, in order to obtain a solution that is 17 % salt. How much of the first solution must be used
I do not know how to set this up. Can you help me please?

Found 2 solutions by ewatrrr, josmiceli:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
set up ...
.11x + .20(4oz) = .17(4oz + x)
.03(4oz)/.06 = x

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +x+ = ounces of 11% solution needed
+.11x+ = ounces of salt in 11% solution
+.2%2A4+=+.8+ ounces of salt in 20% solution
------------------
++%28+.11x+%2B+.8+%29+%2F+%28+x+%2B+4+%29+=+.17+
+.11x+%2B+.8+=+.17%2A%28+x+%2B+4+%29+
+.11x+%2B+.8+=+.17x+%2B+.68+
+.06x+=+.12+
+x+=+2+
2 ounces of 11% solution are needed
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check:
++%28+.11x+%2B+.8+%29+%2F+%28+x+%2B+4+%29+=+.17+
++%28+.11%2A2+%2B+.8+%29+%2F+%28+2+%2B+4+%29+=+.17+
+%28+.22+%2B+.8+%29+%2F+6+=+.17+
+1.02+=+6%2A.17+
+1.02+=+1.02+
OK