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.

Multiply both sides by 100.


Divide both sides by 50.


The amount invested at 7% is
.
The amount invested at 8% is

The amount invested at 9% is

Thus, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.
Are these 2 different questions or is this one equation
Answer:
![-8\left[\begin{array}{ccc}-10&-2&-5\\9&-7&-1\\9&-4&2\end{array}\right]](https://tex.z-dn.net/?f=-8%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-10%26-2%26-5%5C%5C9%26-7%26-1%5C%5C9%26-4%262%5Cend%7Barray%7D%5Cright%5D)
![-\left[\begin{array}{ccc}-80&-56&-40\\72&-56&-8\\72&-32&16\end{array}\right]](https://tex.z-dn.net/?f=-%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-80%26-56%26-40%5C%5C72%26-56%26-8%5C%5C72%26-32%2616%5Cend%7Barray%7D%5Cright%5D)
![=\left[\begin{array}{ccc}56&50&-14\\-120&92&62\\-78&26&2\end{array}\right]](https://tex.z-dn.net/?f=%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D56%2650%26-14%5C%5C-120%2692%2662%5C%5C-78%2626%262%5Cend%7Barray%7D%5Cright%5D)

<u>---------------------------</u>
<u>hope it helps..</u>
<u>have a great day!!</u>
1/2*(-1/7) = - (1*1)/(2*7)= - 1/14
(+)*(-) = (-)
Answer : - 1/14