SOLUTION: I was hoping someone could help me with this Question and show me how you put it in an equation. I'm fine until it gets put in a word problem? I'd like to practice to see what I'm

Question 277598: I was hoping someone could help me with this Question and show me how you put it in an equation. I'm fine until it gets put in a word problem? I'd like to practice to see what I'm doing wrong.
Thanks!!
A person invests $36,000, partly at 5%, partly at 6%, and the remainder at 6.5%. The total annual interest is $2,190. Three times the amount invested at 6% equals the amount invested at 5% and 6.5% combined. How much is invested at each rate?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A person invests $36,000, partly at 5%, partly at 6%, and the remainder at 6.5%.
The total annual interest is $2,190.
Three times the amount invested at 6% equals the amount invested at 5% and 6.5% combined.
How much is invested at each rate?
:
Let a = amt at 5%
Let b = amt at 6%
let c - amt at 6.5%
:
Write an equation for each statement:
:
"A person invests $36,000,"
a + b + c = 36000
:
"The total annual interest is $2,190."
.05a + .06b + .065c = 2190
:
"Three times the amount invested at 6% equals the amount invested at 5% and 6.5% combined."
3b = a + c
:
How much is invested at each rate?
:
We can replace a + c in the 1st equation with 3b
3b + b = 36000
4b = 36000
b =
b = $9000 invested at 6%
:
We know that:
a + c = 3(9000)
a + c = 27000
therefore:
a = (27000-c)
:
In the 2nd (interest) equation replace a with (27000-c), replace b with 9000, find c
.05(27000-c) + .06(9000) + .065c = 2190
1350 - .05c + 540 + .065c = 2190
.05c + .065c = 2190 - 1350 - 540
.015c = 300
c =
c = $20000 invest at 6.5%
:
I'll let you find the amt invested at 5% (a)
:
Check the total interest then

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