You invested $6000 between two accounts paying 3% and 9% annual interest, respectively. If the total interest earned for the year was $480, how much was invested at each rate?

$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)

you would have to sell 24 bracelets.

step-by-step explanation:

84/3.5 = 24