Answer
$\$12,500$ is invested at the $7\%$ rate and $\$7500$ is invested at the $9\%$ rate.
Work Step by Step
We are given that $\$20,000$ is invested in two accounts, at the rate of $7\%$ and $9\%$.
Total yearly interest earned is $\$1550$.
Let the amount x be invested at a rate of $9\%$.
This means that the amount of $\left( 20,000-x \right)$ is invested at a rate of 7%.
Now, it is given that the interest earned in one year is 1550. This implies:
$\left\{ \left( 9 \right)\times \left( x \right) \right\}+\left\{ \left( 7 \right)\times \left( 20,000-x \right) \right\}=\$1550$
This means:
$0.09x+0.07\left( 20,000-x \right)=1550$
Using distributive property:
$0.09x+1400-0.07x=1550$
Now,
$0.02x+1400=1550$
$0.02x=150$
$x=7500$
Thus, amount invested at the rate of $9\%$ is $\$7500$ and the amount invested at $7\%$ is $20,000-x=20000-7500=12500$.
