teresa invested $7600 in two different accounts. the first account paid 6% the second account paid 2% interest. At the end of the first year she had $360 in interest. How much was in each account?

Answer 1

.06x+ .02(7600-x) = 360

.06x- .02x = 360 - (.02)(7600)

.04x = 360- 152 = 208

x = 206/.04 = 25(208) = $5200 invested at 6%

7600-5150 = $2400 invested at 2%

check the answers:

360/7600 = 36/760 = 18/380 = 9/190 = about .047 = about 4.7% average return

6-4.7 = 1.3

4.7-2 = 2.7

2.7/1.3 =about 2.1 =

about 5200/2400= 52/24 = about 2.2

close enough

Answer 2

6% * x + 2% * (7600-x) = 360

4% * x + 152 = 360

.04*x = 208

x = $5,200 at 6% and $2,400 at 2%

Answer 3

Let x be the amount invested at 6%. At the end of the first year:

x(1.06) + (7600)(1.02) - 7600 is the earned interest or

x(.06) + (7600-x)(.02) = 360 (the earned interest)

Solve for x after distributing, grouping terms.