SOLUTION: Geoff has $5000 to invest. He wants to earn $568 in interest in 1 year. He will invest part ot the money at 12% and the other part at 10%. How will he invest to earn $568?

Algebra ->  Finance -> SOLUTION: Geoff has $5000 to invest. He wants to earn $568 in interest in 1 year. He will invest part ot the money at 12% and the other part at 10%. How will he invest to earn $568?      Log On


   



Question 250440: Geoff has $5000 to invest. He wants to earn $568 in interest in 1 year. He will invest part ot the money at 12% and the other part at 10%. How will he invest to earn $568?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
x + y = 5000
x * .12 + y * .10 = 568

use first equation to solve for y in terms of x.

y = 5000 - x

substitute for y in second equation to get:

x * .12 + (5000-x) * .10 = 568

simplify by removing parentheses to get:

x * .12 + 5000*.10 - x*.10 = 568

combine like terms and simplify further to get:

.02*x + 500 = 568

subtract 500 from both sides to get:

.02*x = 68

divide both sides by .02 to get:

x = 68/.02 = 3400.

if x = 3400, then y = 5000 - 3400 = 1600.

.12 * 3400 + .10 * 1600 = 408 + 160 = 568.

his confirms the answer is good.

the answer is:

geoff invests 3400 at 12% and 1600 at 10% to gain 568 interest in one year.