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Last year Sarah invested money in two accounts. The first account had an interest rate of 4%4% and the second account had an interest rate of 3%3%. If she invested $700$⁢700 more in the first account than the second and her total interest income was $392$⁢392, how much did she invest at each rate?

Question

Last year Sarah invested money in two accounts. The first account had an interest rate of 4%4% and the second account had an interest rate of 3%3%. If she invested 700700⁢700 more in the first account than the second and her total interest income was 392392⁢392, how much did she invest at each rate?

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Solution

Let's denote the amount of money Sarah invested in the second account as X. Therefore, the amount of money she invested in the first account is X + $700.

Since the total interest income was $392, we can set up the following equation:

0.04 * (X + 700)+0.03X=700) + 0.03 * X = 392

This equation represents the sum of the interest from both accounts. The 0.04 and 0.03 are the interest rates of the first and second account respectively.

Now, let's solve this equation:

0.04X + 28+0.03X=28 + 0.03X = 392 0.07X + 28=28 = 392 0.07X = 392392 - 28 0.07X = 364X=364 X = 364 / 0.07 X = $5200

So, Sarah invested 5200inthesecondaccount.Andforthefirstaccount,its5200 in the second account. And for the first account, it's 5200 + 700=700 = 5900.

Therefore, Sarah invested 5900at45900 at 4% and 5200 at 3%.

This problem has been solved

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