SOLUTION: A man invests his savings in two accounts, one paying 6% and the other paying 10% simple interest per year. He puts twice as much in the lower-yielding account because it is less r

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A man invests his savings in two accounts, one paying 6% and the other paying 10% simple interest per year. He puts twice as much in the lower-yielding account because it is less r      Log On


   

Question 622619: A man invests his savings in two accounts, one paying 6% and the other paying 10% simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is $4,180. How much did he invest at each rate?
Answer by math-vortex(648) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, there--
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We can use algebra to solve this.
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Let x be the amount of money invested at 6% simple interest.
Let y be the amount of money invested at 10% simple interest.
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Express the interest rates in decimal form: 6% = 0.06 and 10% = 0.10
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The amount of interest earned on the the first account is the interest rate times the amount invested, or 0.06x.
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Likewise, the amount of interest earned on the second account is its interest rate times the amount invested, or 0.10*y.
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Now we need to write two equations using the information in the problem to model the situation.
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The man puts twice as much in the lower-yielding account because it is less risky. In other words,
[the amount invested at 6%] = [2] * [amount invested at 10%]
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In algebra, we can write this relationship as
x=2y
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The amount the man earns in interest is $4,180. In other words,
[interest earned at 6%] + [interest earned at 10%] = [$4,180]
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In algebra, we can write
0.06x%2B0.10y=4180
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We have a system of two equations and two variables. We will use the substitution method to solve for x and y. The first equation states that x=2y, so we substitute 2y for x in the second equation.
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0.06x%2B0.10y=4180
0.06%282y%29%2B0.10y=4180
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Now simplify and solve for x.
0.12y%2B0.10y=4180
0.22y=4180
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Divide both sides of the equation by 0.22
y=19000
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In the context of this problem, y=19000 means that the man invested $19,000 in the account earning 10% simple interest. Since he invested twice as much in the account eating 6%, he must have invested $38,000 at 6%.
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The final step is to check that this allocation of the money actually earns $4,180 in total interest.
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$38,000 at 6% interest earns $2,280 since 0.06*38000=2280
$19,000 invested at 10% interest earns $1,900 since 0.10*19000=1900
$2,280+1,900 is $4,180 so everything checks out.
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That's it. Feel free to email if you have questions about any part of this explanation.
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Ms.Figgy
math.in.the.vortex@gmail.com