SOLUTION: A person invested $6,700 for one year, part at 8%, part at 10% and the remainder at 12%. The total annual income from these investments was $716. The amount of money invested at

Question 82665This question is from textbook Introductory & Intermediate Algebra for College Students
: A person invested $6,700 for one year, part at 8%, part at 10% and the remainder at 12%. The total annual income from these investments was $716. The amount of money invested at 12% was $300 more than the amount invested at 8% and 10% combined. Find the amount invested at each rate. This question is from textbook Introductory & Intermediate Algebra for College Students

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A person invested $6,700 for one year, part at 8%, part at 10% and the remainder at 12%. The total annual income from these investments was $716. The amount of money invested at 12% was $300 more than the amount invested at 8% and 10% combined. Find the amount invested at each rate.
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Let the amt at 8% be x; at 10% be y; at 12% be z:
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Then:
x + y + z = 6700
0.08x + 0.10y + 0.12z = 716
x + y - z = -300
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Solution:
x= $1200
y= $2000
z= $3500
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Cheers,
Stan H.
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