SOLUTION: Stan invested $5,000, part at 8% and part at 17%. If the total interest at the end of the year is $490, how much did he invest at 8%?

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Question 119770: Stan invested $5,000, part at 8% and part at 17%. If the total interest at the end of the year is $490, how much did he invest at 8%?

Answer by Caallen2(9) About Me  (Show Source):
You can put this solution on YOUR website!
You must form your two variables and two equations.
First we know that the total amount is 5000.
Let the 8% portion of the money invested = x
Let the 17% portion of the money invested = y.
The first equation would be x + y = 5000, because each portion added together will be 5000.
The second equation is derived by the percentages of each portion that will equal the interest acquired which is 490. So the second equation would be:
.08x + .17y = 490
Now we have two equations:
x + y = 5000 and .08x + .17y = 490
Solve by substitution.
x + y = 5000
subtract y from both sides
x = 5000 - y
Now substitute 5000 - y for x in the second equation
.08x + .17y = 490
.08(5000 - y) +.17y = 490
Use distributive property
400 - .08y + .17y = 490
Subtract 400 from both sides and simplify
.09y = 90
divide both sides by .09
y = 1000
now substitute the solution for y into the first equation
x = 5000 - y
x = 5000 - 1000
x = 4000
so your answer will be $4000 was invested at 8% and $1000 was invested at 17%.