SOLUTION: Having a real hard time with this...please help! Solve system of equation word problem but either addition method or substitution method. George has two investments that yiel

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Question 113156This question is from textbook
: Having a real hard time with this...please help!
Solve system of equation word problem but either addition method or substitution method.
George has two investments that yield a total of $185.60 in annual interest. The amount invested at 8% is $320 less than twice the amount invested at 6%. How much is invested at each rate?
Amount invested at 8% _________
Amount invested at 6% _________ This question is from textbook

Answer by SHUgrad05(58) (Show Source):
You can put this solution on YOUR website!
To set up this word problem,it's good to start writing out what we know:
We know George has two investments (let's call them x & y), one at 6% and the other at 8%.
The two investments yield a total of $185.60 in annual interest.
So, 0.06x+0.08y=185.60
Next:
We know the amount invested at 8% is $320 less than twice the amount invested at 6% which means: y=2x-320.
So that leaves us with: 0.06x+0.08y=185.60 and y=2x-320
Use the substitution method to continue solving.
0.06x+0.08(2x-320)=185.60
0.06x+0.16x-25.60=185.60
0.22x=211.20
x=$960.00
0.06(960)+0.08y=185.60
57.60+0.08y=185.60
0.08y=128.00
y=$1600
Amt invested at 6%=$960.00
Amt invested at 8%=$1600.00