Systems of equations

Question

Keegan invests $6200 in two different accounts. The first account is paid 6%, the second account paid 2% in interest. At the end of the first year he had earned $324 in interest. How much was in each account?

Answer 1

set up 2 equations

let x = amount invested at 6%

let y = amount invested at 2%

x + y = 6200

.06x + .02 y = 324

solve first equation for y

y = 6200 -x

substitute this into the 2nd equation

.06x + .02(6200 - x) = 324

.06x + 124 - .02x = 324

.06 x - .02x = 200

.04 x = 200

x = 5000

$5000 was invested at 6%

subtract 5000 from 6200 to find that

1200 was invested at 2%