Melissa invested in two accounts. She invested in an account that gave her an interest rate of 4% per year and an account that gave her an interest rate of 10% per year

Question

Melissa invested in two accounts. She invested in an account that gave her an interest rate of 4% per year and an account that gave her an interest rate of 10% per year. She invests a total of $9400, and her interest earnings after one year are $802. Write an equation for the total investment and another equation for the interest earned, and then solve the system to determine how much Melissa invested in each account.let x represent the amount invested with an interest rate of 4% per year and y represent the amount invested with an interest rate of 10% per year.

The equation that describes the total investment is  

The equation that describes the interest earned is   

Amount invested at 4% interest is $ 

Amount invested at 10% interest is $ 

Answer 1

.10x + .4(9400-x) = 802

.10x -.04x = 802 - .04(9400)

.06x = 802 - 802-376 = 426

x = 426/.06 = $7100 invested at 10%

9400-7100 = $2,300 invested at 4%

.1(7.1K) +.04(2.3K) = .71K + .092K = .802K = 802 (K=$1000)

802/9400 = ,08532 = about 8.532%

10 is within 1.468 of 8.532

4 is within 4.532 of 8.532

4.532/1.468 = 3.087

7100/2300 = 3.087