Help with algebra??

Question

Angus invested $18,000, part at 3% and part at 5%. If the total interest at the end of the year is $660, how much did he invest at each rate?
Algebra problem?

I know I have asked for a lot of algebra help, I have only one more week of my course to take and am just trying to get through it. This has been hard for me. Please don't flame me for asking for help. and thank you to those who do help.

Answer 1

let x = the amount you invested at 3% = 0.03
let y = the amount you invested at 5% = 0.05

you are told that you have $18,000 total
and the total interest is $660

that means:
x + y = 18,000
(0.03)x + (0.05)y = 660

use the first equation and solve for x:
x + y = 18,000
x = 18,000 - y

plug x into the 2nd equation:
(0.03)x + (0.05)y = 660
(0.03)(18,000 - y) + (0.05)y = 660

distribute the 3%:
540 - 0.03y +0.05y = 660

simplify:
540 + 0.02y = 660

subtract 540 from each side:
0.02y = 660 - 540 = 120

divide each side by 0.02"
y = (120)/(0.02) = 6000

now sustitute y into the first equation:
x + y = 18,000
x + 6,000 = 18,000

subtract 6,000 from each side:
x = 18,000 - 6,000
x= 12,000

Thus:
x = $12,000
y = $6,000

Answer 2

x + y = 18,000
0.03x + 0.05y = 660

Solving the system of equations you'll get

$ 6,000 invested at 5%
$12,000 invested at 3%

Answer 3

12,000 at 3% = $360
6,000 at 5% = $300

Answer 4

$12,000 invested @ 3%

%6,000 invested @ 5%

Answer 5

wish I could help you sorry