# $6000 are partly invested at 5% and partly at at 9%. If interest earned in a year is $400, what are amounts invested at 5% and at 9%?

Amount invested at 5% is $3,500 and amount invested at 9% is $2,500
Let the amount invested at 5% be $x$, then amount invested at 9% will be $6000 - x$.
Therefore, interest earned on 5% is $\frac{5 x}{100}$ and interest earned on 9% is $\left(6000 - x\right) \frac{9}{100}$. As total interest is $400, we have $\frac{5 x}{100} + \frac{9}{100} \left(6000 - x\right) = 400$or $5 x + 9 \left(6000 - x\right) = 40000$or $5 x + 54000 - 9 x = 40000$or $- 4 x = - 14000$$x = \frac{- 14000}{- 4} = 3500$Hence amount invested at 5% is $3,500 and amount invested at 9% is 6000-3500=\$2,500