Homework Help: algebra

Posted by matt on Monday, November 1, 2010 at 4:54pm.

two investments were made totaling $15,000. for a certain year these investments yielded $1432 in simple interest. part of the $15,000 was invested at 9% and part at 10%. find the amount invested at each rate.

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?

algebra - Reiny, Monday, November 1, 2010 at 6:44pm
first one:
let the amount invested at 9% be x
let the amount invested at 10% be 15000-x

.09x + .10(15000-x) = 1432
multiply each term by 100
9x + 10(15000-x) = 143200
-x = -6800
x = 6800

so $6800 were invested at 9% and $8200 at 10%

do the second question the same way.
algebra - Anonymous, Wednesday, September 7, 2011 at 4:02pm
90+10(15000-x)=1432000

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