SOLUTION: I need your help solving this,$9500 is invested, part at 11% and the other is at 6%. for a certain year the total was $65.00

Algebra ->  Inequalities -> SOLUTION: I need your help solving this,$9500 is invested, part at 11% and the other is at 6%. for a certain year the total was $65.00      Log On


   

Question 590905: I need your help solving this,$9500 is invested, part at 11% and the other is at 6%. for a certain year the total was $65.00
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = amount invest @ 11%
Let +b+ = amount invested @ 6%
given:
(1) +a+%2B+b+=+9500+
(2) +.11a+%2B+.06b+=+65+
-------------------
(2) +11a+%2B+6b+=+6500+
Multiply both sides of (1) by 6 and
subtract (2) from (1)
(1) +6a+%2B+6b+=+57000+
(2) +-11a+-+6b+=+-6500+
+-5a+=+50500+
+a+=+-10100+
The minus sign means it is not possible
Suppose all the money was invested at 6%
+.06%2A9500+=+570+
That's more than $65
I bet the $65 should be $650
Check it