SOLUTION: Suppose you invest money in two accounts. One of the accounts pay 10% annual interest, whereas the other pays 12% annual interest. If you have $3,000 more invested at 12% than

Question 908710: Suppose you invest money in two accounts. One of the accounts pay 10% annual interest, whereas the other pays 12% annual interest. If you have $3,000 more invested at 12% than you invested at 10% , how much do you have invested in each account if the total amount of interest you earn in a year is $1,240 ?

Your investing $ in the 10% account and $ in the 12% account?
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
We don't know the total invested.
We know the difference in the two accounts.
We know that the interest for the two accounts is $1280
0.12*x+0.1*y=1280
We know that the account at 12% has $3000 more.
x=3000+y
We substitute for x
0.12*(3000+y)+0.1*y=1280
We multiply out
360+0.12y+0.1*y=1280
We combine like terms.
0.22*y=920
Isolate y
y=$4181.82 at 10%
x=3000+y
Calculate x
x=$7181.82 at 12%
Now,we know the total invested is: 11363.64
Total invested $7181.82+$4181.82=$11363.64
We check
0.12*7181.82+0.1*4181.82=1280
861.82+418.18=1280
1280.0=1280
Since this statement is TRUE and neither x nor y is negative all is well.

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