SOLUTION: A total of $1150 was investe3d , part of it at 12% and part at 11%. The total yield was $133.75. How much was invested at each rate?

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Question 61075This question is from textbook Algebra 2 with trigonometry
: A total of $1150 was investe3d , part of it at 12% and part at 11%. The total yield was $133.75. How much was invested at each rate? This question is from textbook Algebra 2 with trigonometry

Answer by joyofmath(189) (Show Source):
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A total of $1150 was invested, part of it at 12% and part at 11%. The total yield was $133.75. How much was invested at each rate?
Let A = amount invested at 12%.
Let B = amount invested at 11%.
A total of $1150 was invested so .
The total yield = .
Since , .
Substitute into
and you get .
Simplify and you get .
Simplify more and you get .
Subtract 126.5 from both sides and you get or .
So, $725 was invested at 12% A total of $1150 was invested so the balance, $425, was invested at 11%.
To verify, plug the values of A and B into .
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