SOLUTION: Sam invested $7,500 for one year, part at 12% annual interest and the rest at 10% annual interest. His total interest for the year was $890. How much money did he invest at 12%.

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Sam invested $7,500 for one year, part at 12% annual interest and the rest at 10% annual interest. His total interest for the year was $890. How much money did he invest at 12%.      Log On


   



Question 134468: Sam invested $7,500 for one year, part at 12% annual interest and the rest at 10% annual interest. His total interest for the year was $890. How much money did he invest at 12%.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
x = amount invested at 12%
y = amount invested at 10%
.12x+%2B+.1y+=+890
x+%2B+y+=+7500
x+=+7500+-+y
.12%2A%287500+-+y%29+%2B+.1y+=+890
900+-+.12y+%2B+.1y+=+890
-.02y+=+-10
y+=+10%2F.02
y+=+1000%2F2
y+=+500
x+=+7500+-+y
x+=+7000
He invested $7000 at 12%
check:
.12%2A%287500+-+y%29+%2B+.1y+=+890
.12%2A%287500+-+500%29+%2B+.1%2A500+=+890
.12%2A7000+%2B+50+=+890
840+%2B+50+=+890
890+=+890
OK