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a person had $14,000 infested in two accounts, one paying 9% simple interest and one paying 10% simple interest. how much was invested in each account if the interest at the one year is $1339?

User Andressa
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1 Answer

20 votes
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Given:

a.) A person had 14,000 infested in two accounts.

b.) One paying 9% simple interest.

c.) One paying 10% simple interest.

Let,

x = the amount invested at 9% simple interest

y = the amount invested at 10% simple interest

1.) We know the total amount of money invested is $14,000. We get,

x + y = 14,000

2.) We know that the total interest for the year for the two accounts is $1432. We get,

0.09*x + 0.1*y = 1,339

Let's equate the two equations,

x = 14,000 - y (Substitute for x)

0.09*(14,000 - y) + 0.1*y = 1,339

1,260 - 0.09y + 0.1y = 1,339

0.1y - 0.09y = 1,339 - 1,260

0.01y = 79

0.01y/0.01 = 79/0.01

y = 7,900

Therefore, $7,900 was invested at the rate of 10% simple interest.

Let's determine x, substituting y = 7,900 in x + y = 14,000.

x + y = 14,000

x + 7,900 = 14,000

x = 14,000 - 7,900

x = 6,100

Therefore, $6,100 was invested at the rate of 9% simple interest.

User Polsonby
by
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