Algebra Problem

Question

Larry invested part of his 16,000 at 5% annual simple interest and rest at 6%.  If the total from both accounts was $810 find the amount invested at each rate.

Answer 1

1) x + y = 16,000 or x = 16,000 - y

2) 0.05x + 0.06y = 810

substitute eq(1) into eq(2):

0.05(16,000 - y) + 0.06y = 810

800 - 0.05y + 0.06y = 810

0.01y = 10

y = 10/0.01 = $1,000

using eq(1):

x = 16,000 - 1000 = $15,000

He invested $15,000 at 5% and $1,000 at 6%

check using eq(2):

0.05(15,000) + 0.06(1,000) = 810

750 + 60 = 810