HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!

Question

1) Suppose you invest money in two accounts. One of the accounts pays 4% interest annually, and the other account pays 5% interest annually. You have $ 2000 more invested in the account paying 4% than in the account paying 5%. How much do you have invested in the account paying 4% if you earn $ 670 interest in a year?

2) Money is invested in two accounts, twice as much in one paying 3% interest per year as in the other, which pays 4% interest per year. The total interest earned after one year is $140. If x is the amount invested at the 3% rate and the y is the amount invested at the 4% rate, then write a system of equations that will have a solution that gives the amount invested at each rate, and find the x and y

Answer 1

.05X + .04(X+2) = .67 measured with units of $1000

.09X = .67-.08 = .59

X = .59/.09 = 6.555...= $6,555.56 invested at 5%

X+2 = 8.555 = $8,555.55 invested at 4%

total invested = $15,111.11

5% of 6 5/9 = .05(59/9) = 59/180 = .32777...

4% of 8 5/9 = .04(77/9) = 77/225 = .34222.

.32777 + .34222 = .67 = $670

670/15,111.111 = 4.43% return, slightly less than the average of 4 and 5% because more was invested at 4%

.03(2x) + .04(x) = 140

.06x +.04x = 140

.1x = 140

x = 140/.1 = $1400 invested at 4%

2x = $2800 invested at 3%

.04(1.4) +.03(2.8) using units of $1000

= .056 + .084 = .140 = $140

Answer 2

0.04(x + 2000) + 0.05x = 670

x = 2y

0.03x + 0.04y = 140