use systems of equations to solve?

Question

this will be on investments
my family invests money in two accounts, one of which pays 12% per year, while the other pays 15% per yr. If their total investment is $12,000 and the interest after 1yr is $ 1,650, how much is invested in each account.

Answer 1

Two accounts. One contains 'x' amount of money. The other contains 'y' amount of money, such that

(1) x + y = 12000

After one year, the first account will accrue interest equal to (0.12x), while the second will accrue interest equal to (0.15y). Combined they will total

(2) 0.12x + 0.15y = 1650

Now solve for either x or y. In this case, it is probably easiest to re-write equation (1) as

(3) x = 12000 - y

and plug that into equation (2).

(4) 0.12(12000 - y) + 0.15y = 1650

(5) 1440 - 0.12y + 0.15y = 1650

(6) 0.03y = 210

(7) y = 7000

Now plug that back into equation (3)

(8) x = 12000 - 7000 = 5000

5000 in one account, 7000 in the other. You can work backwards to double-check. 5000 times 12% interest will get you 600 after one year. 7000 times 15% interest will get you 1050 after one year. Combined, 600+1050 equals 1650 total interest after one year.

Answer 2

Let x = amount invested in the first account. Then the amount invested in the second account is 12000 - x.

The interest equation is:

.12x + (12000 - x)*.15 = 1650

.12x + 1800 - .15x = 1650

.03x = 150

x = $5000
second account = $7000

Answer 3

x= investment at 12%
y= investment at 15%

.12(x)+.15(y)=1,650
multiply by 8 1/3 to get
x+1.25(y)=13,750

[x+1.25(y)=13,750] minus [x+y=12,000] equals [.25y=1,750]

y=$7,000
x=12,000-7,000=$5,000