Answer:
$1,000 and $5,000
Step-by-step explanation:
let the 2 amounts be x and y (which add to $6,000)
i.e,
x + y = 6000 -----------(eq1)
assume x is invested at 3% and y is invested at 9%
Assuming simple interest, the following formula is applicable.
I = Prt
where
P = principal ( i.e either $x or $y)
r = rate (i.e 3% or 9%)
t = time elapsed (given as 1 year)
since t = 1 year, the equation reduces to
I = Pr
For $x invested at 3% (i.e 0.03) ---> I = 0.03x
For $y invested at 9% (i.e 0.09) ---> I = 0.09y
given that the total interest for the year = $480,
0.03x + 0.09y = 480 (multiplying both sides by 100)
3x + 9y = 48,000 ---------(eq2)
solving the system of equations made up of eq 1 and eq 2 by elimination,
eq 1 x 3
3x + 3y = 18,000 ------(eq 3)
eq2 - eq3
9y - 3y = 48,000 - 18,000
6y = 30,000
y = 30,00 / 6 = $5,000 (answer)
substitute y = 5,000 into eq 1.
x + 5000 = 6000
x = $1,000 (answer)
