Given:
The rate of interest on three accounts are 7%, 8%, 9%.
She has twice as much money invested at 8% as she does in 7%.
She has three times as much at 9% as she has at 7%.
Total interest for the year is $150.
To find:
Amount invested on each rate.
Solution:
Let x be the amount invested at 7%. Then,
The amount invested at 8% = 2x
The amount invested at 9% = 3x
Total interest for the year is $150.
![x* (7)/(100)+2x* (8)/(100)+3x* (9)/(100)=150](https://img.qammunity.org/2022/formulas/mathematics/college/z7qeouiyfpt86aq3lfjcvqll3qcl0lf28o.png)
Multiply both sides by 100.
![7x+16x+27x=15000](https://img.qammunity.org/2022/formulas/mathematics/college/q1m4cw4o3vkqe666bu3bkq7u1j59zkv4bd.png)
![50x=15000](https://img.qammunity.org/2022/formulas/mathematics/college/t79o9qvm9rpv612zh5rku5y8u5expstezg.png)
Divide both sides by 50.
![x=(15000)/(50)](https://img.qammunity.org/2022/formulas/mathematics/college/ldqd691m2dk14n3jhn4hu4a7qgjx7jpv6f.png)
![x=300](https://img.qammunity.org/2022/formulas/mathematics/college/6cve8hnmmveekedtlygpnav0eausgovnsk.png)
The amount invested at 7% is
.
The amount invested at 8% is
![2(300)=600](https://img.qammunity.org/2022/formulas/mathematics/college/pnc0xk2r16bf1ntww3raqhamh636kgwhj2.png)
The amount invested at 9% is
![3(300)=900](https://img.qammunity.org/2022/formulas/mathematics/college/pq8atazzv9ez61501xxfi68otnbzmdonv2.png)
Thus, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.