Asked by MR.BEAN the 2ND

A self-employed contractor nearing retirement made two investments totaling $15,000. In one year, these investments yielded $1023 in simple interest. Part of the money was invested at 6% and the rest at 7.5%. How much was invested at each rate?

Answered by Reiny
amount invested at 6% .... x
amount invested at 7.5% ----- 15000 - x

.06x + .075(15000-x) = 1023

solve for x, etc

notice this equation follows the same thinking as the equation I gave you in your previous post with the mixtures.
Answered by Steve
this is just another mixture problem. How did you set it up?
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