Question 1115269
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Meisha has $25,000 that she wants to invest. She invests it in accounts paying 12%, 7%, 
and 6% simple interest. The account paying 12% is a higher-risk account, so she wants the
amount in that account to be half of the amount she has in the account paying 6% simple
interest. If her annual interest is $1945, how much is invested at each rate?
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        For this problem (and for many other similar problems) there is much more simpler 

        and more straightforward way to solve, which I will show you right now.



<pre>
Let 'x' be the amount invested at 12%.

Then the amount invested at 6% is 2x, 

and the amount invested at 7% is the rest (25000 - x - 2x) = (25000-3x) dollars.


Now write an equation for the total annual interest 

    0.12x + 0.06(2x) + 0.07*(25000-3x) = 1945  dollars.


Simplify and find x

    0.12x + 0.12x + 1750 - 0.21x = 1945,

    0.12x + 0.12x - 0.21x = 1945 - 1750,

             0.03x         =    195

                 x         =    195/0.03 =  6500  dollars.


Thus, $6500 was invested at 12%;  2*6500 = 13000 dollars were invested at 6% and the rest, 
25000-13000-6500 = $5500  were invested at 7%.
</pre>

Solved.


I neither used a system of three equation nor a system of two equations.


One equation in one unknown was enough to find all three unknowns.


It is a focus-pocus, which works successfully in many other similar problems.