HW HELP NEEDED!

Question

Juan invests a total of $18,000 in two accounts. The first account earned a rate of return of 13% (after a year). However, the second account suffered a 8% loss in the same time period. At the end of one year, the total amount of money gained was $765.00. How much was invested into each account?

$   was invested in the account that gained 13% and

$  was invested in the account that lost 8%.

Answer 1

Lets X is the amount that is invested in 13% so the amount invested in -8% will be 18000 - X. We will find the total interest for each investment and put it equal to 765 then solve for x.

0.13 . X - 0 .08 ( 18000 - X ) = 765

0.13 X + 0.08 X = 765 + 1440

0.21 X = 2205

X = 10500 Dollars invested in 13% and 18000 - 10500 = $7500 was invested in -8% interest, meaning 8% loss.

Answer 2

.13x - .08(18000-x)= 765

where x = amount invested at 13%,

18,000-x = amount invested with 8% l0ss

(.13+.08)x = 765+1440= 2205

x= 2205/.21

= $10,500 invested at 13%

18000-x= $7,500 invested at 8% loss

.13(10500)-.08(7500)= 1365-60765

765/18000=85/2000=.0425= 4.25%=average return

13 is within 8.75 0f 4.25

-8 is within 12.25 of 4.25

12.25/8.75 = 49/35 =7/5=1.4

10,500/7500= 21/15=7/5=1.4

Juan had 1.4 times as much invested at the percent that's closer to the average return.

13% is closer to 4.25% than -8% is to 4.25%