**Question: **A total of 12,000 is invested in two funds paying 9% and 11% simple
interest. If the yearly interest is 1,180. How much of the 12,000 is
invested at each rate?

** Solution: **

Let x: Investment in fund for 9 years.

Let y: Investment in fund for 11 years.

Then the above statement can be written in the mathematical form as:

x + y = 12000

0.09x + 0.11y = 1180

Multiply the first equation by -0.09, we get

-0.09x - 0.09y = -1080

0.09x + 0.11y = 1180

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0.02y = 100

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$\Rightarrow$ y = 5000

To find x, substitute y = 5000 in x + y = 12000

x + 5000 = 12000

$\Rightarrow$ x = 12000 - 5000

$\Rightarrow$ x = 7000

Therefore, x = 7000 is the investment for 9 years and y = 5000 is the investment for 11 years.