<PRE><B>Question 908900</B>
$10,637 is invested, part at 13% and the other at 6%. If the interest earned from the amount invested at 13% exceeds the interest earned from the amount invested at 6% by $880.07, How much is invested at each rate?

$10,637 is invested,
part at 13%  . let it be x

and the other at 6%. let it be y

x+y= 10,637

 interest earned from the amount invested at 13%= 0.13x
 interest earned from the amount invested at 6% = 0.06y 

difference =$880.07, 

0.13x-0.06y=880.07
multiply by 100

13x+6y=88007

x+y = 10,637


1	x	+	1	y	=	10637    	.............1	
								
13	x	+	6	y	=	88007    	.............2	
Eliminate	y							
multiply (1)by		-6						
Multiply (2) by		1						
-6	x		-6	y	=	-63822    		
13	x	+	6	y	=	88007    		
Add the two equations								
7	x				=	24185    		
/	7							
x	=	3455      						
plug value of			x	in (1)				
1	x	+	1	y	=	10637    		
3455		+		y	=	10637    		
				y	=	10637    		-3455
				y	=	7182    		
				y	=	7182   

part at 13%  ----3455

and the other at 6%. 7182
 		1
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