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

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

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