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.
Answers:
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
choice I could help 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%
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