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Last year Katie invested money in two accounts. The first account had an interest rate of 4%4% and the second account had an interest rate of 5%5%. If she invested $100$⁢100 more in the first account than the second and her total interest income was $616$⁢616, how much did she invest at each rate?Step 3 of 3 :  Solve the equation found in part 2 for x. Use this information to answer the given word problem.

Question

Last year Katie invested money in two accounts. The first account had an interest rate of 4%4% and the second account had an interest rate of 5%5%. If she invested 100100⁢100 more in the first account than the second and her total interest income was 616616⁢616, how much did she invest at each rate?Step 3 of 3 :  Solve the equation found in part 2 for x. Use this information to answer the given word problem.

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Solution

From the problem, we know that Katie invested 100moreinthefirstaccountthanthesecond.Let′sdenotetheamountsheinvestedinthesecondaccountasx.Therefore,theamountsheinvestedinthefirstaccountisx+100 more in the first account than the second. Let's denote the amount she invested in the second account as x. Therefore, the amount she invested in the first account is x + 100.

We also know that her total interest income was 616.Theinterestfromthefirstaccountis4616. The interest from the first account is 4% of (x + 100), and the interest from the second account is 5% of x. Therefore, we can set up the following equation:

0.04(x + 100)+0.05x=100) + 0.05x = 616

Solving this equation will give us the value of x, which is the amount Katie invested in the second account. The amount she invested in the first account is x + $100.

Let's solve the equation:

0.04x + 4 + 0.05x = 616 0.09x + 4 = 616 0.09x = 612 x = 612 / 0.09 x = $6800

Therefore, Katie invested 6800inthesecondaccountand6800 in the second account and 6800 + 100=100 = 6900 in the first account.

This problem has been solved

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