Question 853703
The annual interest  on a $14000 investment exceeds the interest on a $7000 investment by $672.  The $14000 is invested at a 0.3% higher rate of interest than the $7000 . what is the interest rate if each investment?

Let interest rate of $14000 be x%

and Interest rate for $7000 be y %

According to the first condition

14000 *x% - 7000*y% = 672
multiply by 100
14000x-7000y = 67200
/700
20x-10y=96.................(1)

II condition
x%=y%+0.3%
x=y+0.3
x-y=0.3..................................(2)

solve (1) & (2)

20	x		-10	y	=	96	.............1	
Total value								
1	x		-1	y	=	0.30	.............2	
Eliminate	y							
multiply (1)by		1						
Multiply (2) by		-10						
20.00	x		-10.00	y	=	96.00		
-10.00	x	+	10.00	y	=	-3.00		
Add the two equations								
10.00	x				=	93.00		
/	10.00							
x	=	9.30						
plug value of			x	in (1)				
20.00	x		-10.00	y	=	96.00		
186.00			-10.00	y	=	96.00		
			-10.00	y	=	96.00		-186.00
			-10.00	y	=	-90.00		
				y	=	9.00		
x=	9.30		Interest rate 	on $14000				
y=	9.00		Interest rate 	on $7000				
m.ananth@hotmail.ca