SOLUTION: A person invested ​$8200 for 1​ year, part at 8​%, part at 9​%, and the remainder at 15​%. The total annual income from these investments was ​$

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Question 1124075: A person invested ​$8200 for 1​ year, part at 8​%, part at 9​%, and the remainder at 15​%. The total annual income from these investments was ​$987. The amount of money invested at 15​% was ​$600 more than the amounts invested at 8​% and 9​% combined. Find the amount invested at each rate.
The person invested ​$ nothing at 8​%,
​$nothing at 9​%,
and ​$ nothing at 15​%.

Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.
A person invested ​$8200 for 1​ year, part at 8​%, part at 9​%, and the remainder at 15​%.
The total annual income from these investments was ​$987. The amount of money invested at 15​% was ​$600 more
than the amounts invested at 8​% and 9​% combined. Find the amount invested at each rate.
~~~~~~~~~~~~~~~~~~~~


            At the first glance, this problem is for 3 unknowns and 3 equations.

            But it is really big deal to solve such a system.

            Fortunately, there is a way to reduce the solution to two unknowns and two equations, which is much easier to solve.

            Below I will show you how to do it.


Notice this phrase of the condition

    "The amount of money invested at 15​% was ​$600 more than the amounts invested at 8​% and 9​% combined".


It means that if you subtract $600 from the total sum of $8200, and then divide this difference by 2, 

you will get how much was invested at 8% and 9% combined:   = 3800 dollars.


Thus you just know that the combined amount invested at 8% and 9% is $3800.

Hence, the rest  $8200 - $3800 = $4400  was invested at 15%.


Also, now you can easily calculate interest from $4400 invested at 15%: it is  0.15*4400 = 660 dollars.



   At this point you just, actually reduced the original problem to 2 unknowns and 2 equations.



Let x = the amount invested at 8% and y = the amount invested at 9%.


Then 

    x +     y = 3800       dollars      (1)
0.08x + 0.09y = 987 - 660  dollars      (2)


Do you understand how I obtained the equation (2) and what does it mean ?

It simply means that combined interest from the 8% account and 9% account is equal to the total interest 
of the three accounts minus the interest of the 15% account, which we calculated above.


OK, very good.  Any way, we just reduced the original problem to the standard equations for two investments:

    x +     y = 3800             (1)
0.08x + 0.09y =  327             (2)


To solve it you can use the Substitution method.  From eq(1) express  x = 3800-y
and substitute it into eq(2).  You will get

0.08*(3800-y) + 0.09y = 327

0.08*3800 - 0.08y + 0.09y = 327

0.01y = 327 - 0.08*3800

y =  = (at this point I use Excel spreadsheet in my computer and get the answer in one click) = 2300 dollars.


Thus you got that the amount at 9% is 2300 dollars.

Then the amount at 8% is the rest  3800 - 2300 = 1500 dollars.


Answer.  $1500 at 8%;  $2300 at 9%  and  $4400 at 15%.


Check.  a)  the combined amount invested at 8% and 9% is  1500 + 2300 = 3800 dollars;

            It is $600 dollars less than $4400 invested at 15%.   ! Correct !

        b)  Total investment is  1500 + 2300 + 4400 = 8200 dollars.   ! Correct !

Solved.

-------------------

The lesson to learn from my solution is THIS :

    At the first glance, this problem is for 3 unknowns and 3 equations.

    But you learned, and I showed it to you, how to reduce the problem to 2 equation in 2 unknowns.

-------------------

To see many other similar problems solved by the same method,  look into the lesson
    - HOW TO algebreze and solve this problem on 2 equations in 2 unknowns
in this site.

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