SOLUTION: Mr. Traynor invested a sum of money at 6%. He invested a second sum, $250 more than the first sum, at 8%. If his total annual income was $300, how much did he invest?

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Mr. Traynor invested a sum of money at 6%. He invested a second sum, $250 more than the first sum, at 8%. If his total annual income was $300, how much did he invest?      Log On


   

Question 1103689: Mr. Traynor invested a sum of money at 6%. He invested a second sum, $250 more than the first sum, at 8%. If his total annual income was $300, how much did he invest?
Found 2 solutions by stanbon, richwmiller:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Mr. Traynor invested a sum of money at 6%. He invested a second sum, $250 more than the first sum, at 8%. If his total annual income was $300, how much did he invest?
----
Equation:
int + int = int
0.06x + 0.08(x+250) = 300
------
6x + 8x + 200 = 30000
------------------------
14x = 29800
x = $2128.57 (amt. invested at 6%)
x + 250 = $2378.57 (amt. invested at 8%)
-----------
Cheers,
Stan H.
----------

Answer by richwmiller(17219) About Me  (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 $300
0.08*x+0.06*y=300
We know that the account at 8% has $250 more.
x=250+y
We substitute for x
0.08*(250+y)+0.06*y=300
We multiply out
20+0.08y+0.06*y=300
We combine like terms.
0.14*y=280
Isolate y
y=$2000 at 6%
x=250+y
Calculate x
x=$2250 at 8%
Total invested $2250+$2000=$4250
Now,we know the total invested is: $4250
We check
0.08*2250+0.06*2000=300
180+120=300
300=300
Since this statement is TRUE and neither x nor y is negative all is well.
codeintmt