# Help next to algebra??

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

I know I have asked for a great deal of algebra help, I enjoy only an added week of my course to take and am a moment ago trying to get through it. This have been frozen for me. Please don't flame me for asking for help. and thank you to those who do relieve.

let x = the amount you invested at 3% = 0.03
tolerate 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 funds:
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 respectively 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 respectively side:
x = 18,000 - 6,000
x= 12,000

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

wish I could relieve you sorry

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%

Source(s):

my brains

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

%6,000 invested @ 5%