SOLUTION: Stan invested $17,000, part at 18% and part at 4%. If the total interest at the end of the year is $1,380, how much did he invest at each rate?

Algebra ->  Inequalities -> SOLUTION: Stan invested $17,000, part at 18% and part at 4%. If the total interest at the end of the year is $1,380, how much did he invest at each rate?      Log On


   



Question 52685: Stan invested $17,000, part at 18% and part at 4%. If the total interest at the end of the year is $1,380, how much did he invest at each rate?
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
Let the amount of principle invested at 18%=x
Then the amount invested at 4% would be 17,000-the amount he invested at 18%=17,000-x
time = 1
interest=1,380
Interest = principle*rate*time
1,380=.18x+.04(17,000-x)Multiply both sides by 100 to clear the decimals.
100(1,380)=100(.18x+.04(17,000-x))
100(1,380)=100(.18x)+100(.04)(17,000-x)
138,000=18x+4(17,000-x)
138,000=18x+4(17,000)+4(-x)
138,000=18x+68,000-4x
138,000=(18-4)x+68,000
138,000=14x+68,000
138,000-68,000=14x+68,000-68,000
70,000=14x
70,000/14=14x/14
5,000=x
There was $5,000 invested at 18%
There was 17,000-5,000=$12,000 invested at 4%