Systems of linear equetikn

Question

Button invests $5000 in two different accounts. The first account paid 10%, 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

.10x + .02(5000-x) = 324

.10x -.02x = 324 -.02(5000)

.08x = 324 -100 = 224

x = 224/.08 = amount invested at 10% =$2800

5000-x =5000- 2800= 2200

= amount invested at 2% = $2200

.10(2800) + .02(2200)

=280+44= 324

324/5000 = .0648 = average 6.48% return on $5,000

10-6.48 = 3.52

6.48-2 = 4.48

3.52/4.48 = 88/112=11/14

= 2200/2800 = 11/14

10% is closer to 6.48% so more is invested at 10 than 2%

exactly 14/11 times as much at 10 than at 2%

14/11 x 2200 = 2800