Question 532306
An investor has a total $18,000 deposited in three different accounts,

9%,------------ $x
7%,-------------$y
5%.-------------$z 

The account deposited in the 9% account is twice the amount in the 5% account.
x=2z

x+y+z=18000
substitute z
2z+y+z=18000
y+3z=18000.................1
the three accounts earn total annual interest of $1340, 
9%x+7%y+5%z=1340
x=2z
9%*2z+7%y+5%z=1340
18%z+7%y+5%z=1340
multiply by 100
18z+7y+5z=134000
23z+7y=134000................2


3	z	+	1	y	=	18000	.............1	
23	z	+	7	y	=	134000	.............2	
Eliminate y								
multiply (1)by 		-7						
Multiply (2) by		1						
-21	z		-7	y	=	-126000		
23	z		7	y	=	134000		
Add the two equations								
2	z				=	8000		
/	2							
z	=	4,000.00						
plug value of 			z	in (1)				
3	z	+	1	y	=	18000		
12000		+	1	y	=	18000		
			1	y	=	18000		-12000
			1	y	=	6000		
				y	=	6000.00	

Now that you know z & y you can calculate x 	
m.ananth@hotmail.ca