SOLUTION: 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?

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Question 49844: 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?
Answer by tutorcecilia(2144) About Me  (Show Source):
You can put this solution on YOUR website!
Since this is a linear equation (total amount earned increases with the rate of interest) use the standand formula of a line:
Ax+By=C [Standard Formula for the equation of a line]
A = rate of A = 3% = 0.03
x = amount of x invested = unknown amount
B = rate of B = 5% = 0.05
y = (total amount invested ($18,000) minus x)
C = total amount of interest earned
.
Ax+By=C
(0.03)(x) + (0.05)(18,000-x) = 660 [Plug-in the values]
0.03x + (900-0.05x) = 660 [Simplify]
-0.02x + 900 = 660
-0.02x + 900 = 660 [Solve for x]
x = 12,000 invested
So, $12,000 invested at a rate of 3% = 12,000 x 0.03 = $360 interst earned
.
(18,000-x)= 18,000-12,000= $6,000 invested
So, y = $6,000 invested at 5% = 6,000 x .05 = $300 interest earned
.
Checking:
$12,000 + $6,000 = $18,000 total invested
360 + 300 = $660 total interest earned