SOLUTION: Hello, I'm having trouble with this word problem: Wendell invested a portion of $10,000 at 11% annual interest and the balance at 8% annual interest. He earned $530 more

Question 85253: Hello,
I'm having trouble with this word problem:

Wendell invested a portion of $10,000 at 11% annual interest and the balance at 8% annual interest. He earned $530 more interest for the year from the account earning 11% annual interest than from the account earning 8% annual interest. How much did he invest at 8%?

Thank you!!!
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!

Let x=amount invested at 11%
Then 10,000-x=amount invested at 8%
Interest (I)=Principal(P)*Rate(R)*Time(T) --here T=1
Interest earned at 11%=0.11x
Interest earned at 8%=0.08(10,000-x)
Now we are told that:
0.08(10,000-x)+530=0.11x get rid of parens
800-0.08x+530=0.11x add 0.08x to both sides
800-0.08x+0.08x+530=0.11x+0.08x collect like terms
1330=0.19x divide both sides by 0.19
x=7000---------------------amount invested at 11%
10,000-x=10,000-7000=3000----------------amount invested at 8%

CK
0.08(3000)+530=0.11(7000)
240+530=770
770=770

Hope this helps----ptaylor

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