Question 1130905
<br>
let x = amount invested at 9%
then 4000-x = amount invested at 6%<br>
interest from investment at 9% = .09(x)
interest from investment at 6% = .06(4000-x)<br>
The total interest is $286:<br>
{{{.09(x)+.06(4000-x) = 286}}}<br>
You can do the algebra to finish the problem.<br>
Note that the problem is poorly formulated, because the amounts invested at the two rates are not whole numbers of dollars, or even whole numbers of cents... not realistic!