Linear Inequalities problem solving

Question

You invested $6000 between two accounts paying 2% and 7% annual interest, respectively. If the total interest earned for the year was $220, how much was invested at each rate?

Answer 1

Assuming simple interest.

i = prt

For this case, t = 1

Let x and y be the two invested amounts.

x+y = 6000 ==> y = 6000-x

.02x + .07y = 220

.02x + .07(6000-x) = 220

.02x + 420 - 0.7x = 220

-0.05x = -200

x = -200/-0.05 = 4000

y = 6000-4000 = 2000

$4000 was invested at 2% and $2000 was invested at 7%

Answer 2

.07x + .02(6000-x)= 220

.07x -.02x = 220-120

.05x = 100

x =$2000 invested at 7%

$4,000 invested at 2%

check the answers

220/6000= 11/300 = 3 2/3% average interest rate

7- 3 2/3 = 3 1/3

3 2/3 -2 = 1 2/3

(10/3)/(5/3) = 2 times as much invested at 2%

4,000/2,000= 2

.07(2,000) + .02(4000) = 140+800= 220