SOLUTION: A man invest a certain amount of money at 2% interest and $800 more than that amount in another account at 4% interest. At the end of one year, he earned $92 in interest. How much

Algebra ->  Finance -> SOLUTION: A man invest a certain amount of money at 2% interest and $800 more than that amount in another account at 4% interest. At the end of one year, he earned $92 in interest. How much       Log On


   

Question 993978: A man invest a certain amount of money at 2% interest and $800 more than that amount in another account at 4% interest. At the end of one year, he earned $92 in interest. How much money was invested in each account
Found 2 solutions by Boreal, stanbon:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x=2% amount
x+800=4%
.02x+.04(x+800)=92
.02x+ .04x + 32=92
.06x=60
multiply both by 100
6x=6000
x=1000
x+800=1800
2% of 1000=$20
4% of 1800=$72
$1000 at 2%
$1800 at 4%

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A man invests a certain amount of money at 2% interest and $800 more than that amount in another account at 4% interest. At the end of one year, he earned $92 in interest. How much money was invested in each account
------
Equation:
int + int = 92 dollars
-----
0.02x + 0.04(x+800) = 92
---------
2x + 4x + 3200 = 9200
---------
6x = 6000
x = 1000 (amt. invested at 2%)
----
x+800 = 1800 (amt. invested at 4%)
-------
Cheers,
Stan H.
-----------