SOLUTION: Mary invested a total of ​$48,000 in three different bank accounts. One account pays an annual interest rate of 2​%, the second account pays 3​% annual​ int

Algebra ->  Finance -> SOLUTION: Mary invested a total of ​$48,000 in three different bank accounts. One account pays an annual interest rate of 2​%, the second account pays 3​% annual​ int      Log On


   



Question 1107868: Mary invested a total of ​$48,000 in three different bank accounts. One account pays an annual interest rate of 2​%, the second account pays 3​% annual​ interest, and the third account pays 4​% annual interest. In one​ year, Mary earned a total of ​$1,550 in interest from these three accounts. If Mary invested ​$8,000 more in the account that pays 3​% interest than she did in the account that pays 4​% ​interest, find the amount she invested in each account.
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x is 2%
y=3%
z=4%
but z=y-8000
Therefore,
x+y+y-8000=48000
x+2y=56000
.02x+.03y+.04(y-8000)=1550
.02x+.03y+.04y-320=1550
.02x+.07y=1870
x+2y=56000
multiply the top by -50
-x-3.5y=-93500
add
-1.5y=-37500
y=25000@3%=$750
y-8000=17000@4%=$680
remaining is x
6000@0.02 for $120
invested
$6000 at 2%
$25000 at 3%
$17000 at 4%