Mrs. Ming invested an amount of money in two accounts for one year. She invested some at 8% interest and the rest at 6% interest. Her total amount invested was $1,500. At the end of the year, she had earned $106.40 in interest. How much had Mrs. Ming invested in the account paying 6%?

$680 was invested at 6%, and $820 was invested at 8%.

Step-by-step explanation:

Let's first let x be the amount she invested in the account with 8% interest, and y be the amount invested in the account with 6% interest.

We are given that the total amount invested was 1500, thus:

x + y = 1500

At the end of the year, she earned 106.40 in interest. Interest is taken by multiplying the amount invested and the rate. Time also plays a factor, but since she only invested for one year, and it isn't specified whether simple or compound interest is used, we can just use the interest rate and the amount.

Interest earned at 8% = 0.08x

Interest earned at 6% = 0.06y

Now, from the given: 0.08x + 0.06y = 106.40.

We can multiply this equation by 100 to remove the decimal part:

8x + 6y = 10640

We can substitute, from the earlier x + y = 1500, x = 1500-y:

8(1500 - y) + 6y = 10640

Solving for y,

12000 - 8y + 6y = 10640

-2y = -1360

y = $680

Then since x = 1500-y, x = $820.

To check,

680 + 820 = 1500, correct

820*0.08 + 680*0.06 = 65.6 + 40.8 = 106.4, correct.

Thus, $680 was invested at 6%, and $820 was invested at 8%.

step-by-step explanation:

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