SOLUTION: Please help me with this problem. I do not know how to set it up. How can $5,400 be invested, part of it at 8% and the rest of it invested at 10%, so that the two investments w

Question 164192: Please help me with this problem. I do not know how to set it up.
How can $5,400 be invested, part of it at 8% and the rest of it invested at 10%, so that the two investments will produce the same amount of interest?
Found 2 solutions by alicealc, aka042:
Answer by alicealc(293) (Show Source): You can put this solution on YOUR website!
Let:
first investment = x
second investment = 5400 - x

the two investments has the same amount of interest, so:

<=> 0.08x = 0.1*(5400 - x)
<=> 0.08x = 540 - 0.1x
<=> 0.08x + 0.1x = 540
<=> 0.18x = 540
<=> x = 540/0.18
<=> x = 3000

first investment = 3000
second investment = 5400 - 3000 = 2400

so you should invest $3,000 at 8% and $2,400 at 10%

Answer by aka042(26) (Show Source): You can put this solution on YOUR website!
Let's call the part of money invested at 8% x. Therefore, that amount of money will earn x*.08 interest. The rest of the money will be invested at 10%. The rest of the money is (5400-x), because it is whatever part of the 5400 was not invested at 8% (x). Therefore, that money will earn .1(5400-x) interest. Now we just need to equate these two expressions:

which we can expand to . Let's combine like terms to arrive at . Finally, divide both sides to arrive at . This means 3000 is invested at 8% interest and 5400-3000= 2400 is invested at 10% interest.

To check our answer, we see that 3000 * .08 = 240 and 2400*.1 = 240
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