Last year Rachel invested money in two accounts. The first account had an interest rate of 6%6% and the second account had an interest rate of 4%4%. If she invested $800$800 more in the first account than the second and her total interest income was $238$238, 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 Rachel invested money in two accounts. The first account had an interest rate of 6%6% and the second account had an interest rate of 4%4%. If she invested 800 more in the first account than the second and her total interest income was 238, 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.
Solution
From the problem, we know that Rachel invested 800.
We also know that her total interest income was 800), and the interest from the second account (with a 4% interest rate) would be 0.04 * x.
Setting up the equation based on the total interest income, we get:
0.06 * (x + 238
Expanding and simplifying the equation, we get:
0.06x + 238
Combining like terms, we get:
0.10x + 238
Subtracting $48 from both sides, we get:
0.10x = $190
Finally, dividing both sides by 0.10, we find that x = $1900.
So, Rachel invested 1900 + 2700 in the first account (with a 6% interest rate).
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