solve this question please

Question

Ngoc invests a total of $26,500 in two accounts. The first account earned an annual interest rate of 6% and the second account earned an annual interest of 5%. At the end of one year, the total amount of money gained was $1,485.00. How much was invested into each account?

$ was invested in the account that earned 6% and

$ was invested in the account that earned 5%.

Answer 1

This is an Algebra problem and has nothing at all to do with Trigonometry.

Let X = amount of money invested at 6% interest.

Let Y = amount of money invested at 5% interest.

X + Y = 26,500

.06X + .05Y = 1485 (I don't like using 'decoration zeros' in front of the decimal point when doing calculations.)

Solve simultaneously to get X = 16,000 and Y = 10,500 and don't forget to check your answer.

$16,000 was invested in the account earning 6% interest.

$10,500 was invested in the account earning 5% interest.

Answer 2

Let "f" be the amount he invested in the first account. Let "s" be the amount he invested in the second account.

The f + s = 26500 which means f = 26500 - s

Using the fact that 1 year went by and each account earned a certain percentage interest in a year then we can say:

0.06f + 0.05s = 1485

Plugging in "26500 - s" in place of "f" (because f = 26500 - s), we get:

0.06(26500 - s) + 0.05s = 1485

1590 - 0.06s + 0.05s = 1485

-0.01s = -105

s = 10500

If s = 10500 then f = 26500 - 10500 or f = 16000

So $16,000 in the first account and $10,500 in the second account