Question 768059
A person invested $7800 for one year,part at 6%, part at 9%, and the remainder at 12%.The total annual income from these investments was $795.The amount of money invested a t 12% was $1200 more than the amounts invested at 6% and 9% combined.What was the amount invested at each rate?

6%--------------------x
9%--------------------y

12%-------------------7800-(x+y)

TOTAL INCOME
6%x+9%y+12%*7800-(x+y) = 795

SECOND CONDITION

7800-(x+y)=x+y+1200


6%x+9%y+12%*7800-(x+y) = 795
multiply equation by 100
6x+9y+12*(7800-(x+y)) =79500

6x+9y+93600-12x-12y=79500
-6x-3y=79500-93600
-6x-3y=-14100
/-3
2x+y=4700........................(1)

7800-(x+y)x+y+1200
7800-x-y=x+y+1200
2x+2y=6600
/2
x+y=3300...........................(2)

solve eqn (1) & (2)

2	x	+	1	y	=	4700	.............1	
Total value								
1	x	+	1	y	=	3300	.............2	
Eliminate	y							
multiply (1)by		-1						
Multiply (2) by		1						
-2	x		-1	y	=	-4700		
1	x	+	1	y	=	3300		
Add the two equations								
-1	x				=	-1400		
/	-1							
x	=	1400						
plug value of			x	in (1)				
2	x	+	1	y	=	4700		
2800		+		y	=	4700		
				y	=	4700		-2800
				y	=	1900		
				y	=	1900		
x=	1400investment at6%						
y=	1900 investment at 9%	
Balance at 12%					
m.ananth@hotmail.ca