SOLUTION: I don't know where to start, please help! Albert invests money in three accounts that pay 4%, 10%, and 15%. He has two times as much money invested at 10% than he does at 15%. If

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: I don't know where to start, please help! Albert invests money in three accounts that pay 4%, 10%, and 15%. He has two times as much money invested at 10% than he does at 15%. If       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 838759: I don't know where to start, please help!
Albert invests money in three accounts that pay 4%, 10%, and 15%. He has two times as much money invested at 10% than he does at 15%. If the total amount he has invested is $4,950 and his interest for the year comes to $520, how much money does he have in each account?
I know how to do it when there are only two varialbes, but no clue how to do it where there are three.

Found 2 solutions by Fombitz, josmiceli:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let the three amounts be A, B, and C.
1.B=2C
2.A%2BB%2BC=4950
4%2AA%2F100%2B10%2AB%2F100%2B15%2AC%2F100=520
3.4A%2B10B%2B15C=52000
When you substitute eq. 1 into eqs.2 and 3, you end up with only 2 variables (A & C).
A%2B2C%2BC=4950
4.A%2B3C=4950
.
.
.
4A%2B20C%2B15C=52000
5.4A%2B35C=52000
.
.
Now solve for A and C and go back to eq. 1 to solve for B.
Repost if you still need help.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
There are 3 unknowns, and you can get
3 equations from given data
---------------------------
Let +a+ = amount invested @ 4%
Let +b+ = amount invested @ 10%
Let +c+ = amount invested @ 15%
----------------------------------
(1) +a+%2B+b+%2B+c+=+4950+
(2) +.04a+%2B+.1b+%2B+.15c+=+520+
(3) +b+=+2c+
-------------------------------
(2) +4a+%2B+10b+%2B+15c+=+52000+
and
Multiply both sides of (1) by +4+
and subtract (1) from (2)
(2) +4a+%2B+10b+%2B+15c+=+52000+
(1) +-4a+-+4b+-+4c+=+-19800+
+6b+%2B+11c+=+32200+
Substitute (3) into this result
+6%2A%282c%29+%2B+11c+=+32200+
+23c+=+32200+
+c+=+1400+
and, since
(3) +b+=+2c+
(3) +b+=+2800+
and
(1) +a+%2B+b+%2B+c+=+4950+
(1) +a+%2B+2800+%2B+1400+=+4950+
(1) +a+%2B+4200+=+4950+
(1) +a+=+750+
-----------------
$750 is invested @ 4%
$2800 is invested @ 10%
$1400 is invested @ 15%
----------------------
check:
(2) +.04a+%2B+.1b+%2B+.15c+=+520+
(2) +.04%2A750+%2B+.1%2A2800+%2B+.15%2A1400+=+520+
(2) +30+%2B+280+%2B+210+=+520+
(2) +520+=+520+
OK