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A woman earns $ 1,350 in interest from two accounts in one year. If she has three times as much invested at 7% as she does at 6%, how much does she have in each account?​

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Answer:

The woman invested $15,000 at 7% interest rate and $5,000 at 6% interest rate.

Explanation:

We are given the following in the question:

Let x be the interest earned from 7% interest rate and y be the interest earned from 6% interest rate.

The woman invested has three times as much invested at 7% as she does at 6%.

Thus, we can write the equation:


x = 3y

The total interest is $1,350.

Thus, we can write the equation:


1350 = (7x)/(100) + (6y)/(100)\\\\7x + 6y = 135000

Solving the two equations by substitution method:


7(3y) + 6y = 135000\\27y = 135000\\y = 5000\\x = 3(5000) = 15000

Thus, she invested $15,000 at 7% interest rate and $5,000 at 6% interest rate.

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