You invested $3000 between two accounts paying 6% and 8% annual​ interest, respectively. If the total interest earned for the year was $200, how much was invested at each​ rate?

Answer 1

Let X = amount invested at 6%

Let Y = amount invested at 8%

X + Y = 3000

.06X +.08Y = 200

I will solve using substitution and rewrite the first equation as Y = 3000 - X

.06X + .08(3000 - X) = 200

.06X + 240 - .08X = 200

-.02X = -40

X = 2000

Y = 1000

Now checking, .06( 2000) + .08(1000) =? 200

120 + 80 =? 200

200 = 200

$2000 is invested at 6% annual interest earning $120

$1000 is invested at 8% annual interest earning $80