SOLUTION: You invested $10,000 in two stocks paying 9% and 12% annual interest, respectively. At the end of the year, the total interest from these investments was $930. How much was in

Algebra ->  Linear-equations -> SOLUTION: You invested $10,000 in two stocks paying 9% and 12% annual interest, respectively. At the end of the year, the total interest from these investments was $930. How much was in      Log On


   

Question 79731: You invested $10,000 in two stocks paying 9% and 12% annual interest,
respectively. At the end of the year, the total interest from these investments
was $930. How much was invested at each rate?
Found 2 solutions by checkley75, DarkOnyx:
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
.12x+.09(10000-x)=930
.12x+900-.09x=930
.03x=30
x=30/.03
x=1000 invested @ 12%
10000-1000=9000 invested @ 9%
proof
.12*1000+.09*9000=930
120+810=930
930=930

Answer by DarkOnyx(1) About Me  (Show Source):
You can put this solution on YOUR website!
Let x= the amount invested at 9%
Let 10000-x = the amount invested at 12%
0.09x+ 0.12(10000-x)=930
0.09x+1200-0.12x=930
-0.03x=--270
x=9000
9000 is the amount invested at 9%
Check:
0.09(9000)+0.12(10000-9000)=930
810+120=930
930=930