SOLUTION: Amelia's parents and grandparents have both opened savings accounts to save for Amelia's education. This year, Amelia's parents' savings account earned 5% interest and her grandpar
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Question 179603: Amelia's parents and grandparents have both opened savings accounts to save for Amelia's education. This year, Amelia's parents' savings account earned 5% interest and her grandparent's account earned 8% interest. This year, the two savings accounts have earned a total of $460 in interest. At the beginning of the year, the two accounts were worth a total of $6500. Determine the current value of both saving accounts. Show all work necessary to justify your answer.
I am not sure how to break down this problem in order to get the answer. I realize it is two parts.
a+b=$460
x+y=$6960
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Amelia's parents and grandparents have both opened savings accounts to save for Amelia's education. This year, Amelia's parents' savings account earned 5% interest and her grandparent's account earned 8% interest. This year, the two savings accounts have earned a total of $460 in interest. At the beginning of the year, the two accounts were worth a total of $6500. Determine the current value of both saving accounts. Show all work necessary to justify your answer.
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Equation:
interext + interest = interest
Amelia's parents and grandparents have both opened savings accounts to save for Amelia's education. This year, Amelia's parents' savings account earned 5% interest and her grandparent's account earned 8% interest. This year, the two savings accounts have earned a total of $460 in interest. At the beginning of the year, the two accounts were worth a total of $6500. Determine the current value of both saving accounts. Show all work necessary to justify your answer.
0.05x + 0.08(6500-x) = 460
Multiply thru by 100 to get:
5x + 8(6500-x) = 46000
-3x + 8*6500 = 46000
-3x = -6000
x = $2000 (amt. invested at 5%)
6500-2000 = $4500 (amt. invested at 8%)
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Cheers,
Stan H.
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