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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?
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! 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
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