SOLUTION: Different amounts are invested at 6%, 7%, and 8% yearly interest. The amount invested at 7% is $300 more than what is invested at 6%, and the total yearly income from all three inv

Question 329657: Different amounts are invested at 6%, 7%, and 8% yearly interest. The amount invested at 7% is $300 more than what is invested at 6%, and the total yearly income from all three investments is $208. A total of $2900 is invested. Find the amount invested at each rate.
THANKS TO ANYONE WHO CAN HELP!!!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Different amounts are invested at 6%, 7%, and 8% yearly interest.
The amount invested at 7% is $300 more than what is invested at 6%,
and the total yearly income from all three investments is $208.
A total of $2900 is invested.
Find the amount invested at each rate.
:
Let x = amt invested at 6$
Let y = amt invested at 7%
Let z = amt invested at 8%
:
Write an equation for each statement:
:
"The amount invested at 7% is $300 more than what is invested at 6%,"
y = x + 300
:
"the total yearly income from all three investments is $208."
.06x + .07y + .08z = 208
get rid of the decimals, mult by 100
6x + 7y + 8z = 20800
Replace y with (x+300)
6x + 7(x+300) + 8z = 20800
6x + 7x + 2100 + 8z = 20800
13x + 8z = 20800 - 2100
13x + 8z = 18700
:
"A total of $2900 is invested."
x + y + z = 2900
Replace y with (x+300)
x + (x+300) + z = 2900
2x + z = 2900 - 300
2x + z = 2600
:
Two equation two unknowns, use elimination here,
mult the above equation by 8, subtract: 13x + 8z = 18700
16x + 8z = 20800
13x + 8z = 18700
------------------subtraction eliminates z, find x
3x = 2100
x =
x = $700 invested at 6%
:
Find y using, y = x + 300
y = 700 + 300
y =$1000 invested at 7%
:
Find z using, x + y + z = 2900
700 + 1000 + z = 2900
z = 2900 - 1700
z = $1200 invested at 8%
:
:
Check the solution by finding the total interest
.06(700) + .07(1000) + .08(1200) =
42 + 70 + 96 = 208; confirms our solutions of x=700, y=100, z=1200

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