Question

Please help!
Joe invested part of $8000 at 11% annual interest. The rest was invested at 9%,last years interest on the $8000 was $850. How much did he invest at each rate?

Answer 1

let 1 value be X and the other be Y

X + Y = 8000 --1
0.11X + 0.09Y = 850 --2
0.09X + 0.09Y = 8000*0.09 = 720 -- 3 x 0.09 into 1
0.11X + 0.09Y = 850 --2
-0.02X = -130
X = 6500
Y = 8000 - 6500 = 1500