Question 1044420

You invested ​$29,000 in two accounts paying 6% and 9% annual​ interest, respectively.
If the total interest earned for the year was $1980, how much was invested at each​ rate? 
<pre>Let amount invested at 6% be S
Then amount invested at 9% = $29,000 - S
The following INTEREST equation is then formed: .06S + .09(29,000 - S) = 1,980
.06S + 2,610 - .09S = 1,980
.06S - .09S = 1,980 - 2,610
- .03S = - 630
S, or amount invested at 6% = {{{(- 630)/(- .03)}}}, or {{{highlight_green("$21,000")}}}
Amount invested at 9%: $29,000 - $21,000, or {{{highlight_green("8,000")}}}
It is that simple!