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6,000 dollars is invested in two different accounts earning 3%and 5% interest. At the end of the year, the two accounts earned 220 in interest. How much money was invested at 3%…
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6,000 dollars is invested in two different accounts earning 3%and 5% interest. At the end of the year, the two accounts earned 220 in interest. How much money was invested at 3%…
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Oct 27, 2018
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6,000 dollars is invested in two different accounts earning 3%and 5% interest. At the end of the year, the two accounts earned 220 in interest. How much money was invested at 3%
Mathematics
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Lanenok
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Hi there
To solve this question
0.03x+0.05 (6000-x)=220
Solve for x to know How much money was invested at 3%
First multiply the values
0.03x+0.05×6000-0.05x=220
0.03x+300-0.05x=220
Second calculate the difference and rearranging the values
0.03x-0.05x=220-300
-0.02x=-80
Divide both sides by-0.02 and divide the values
X=-0.02/-80
x=4000 invested at 3% this is the answer of your question
While the amount invested at 5% is
6000-4000=2000
Good luck!
RafaelGP
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Oct 31, 2018
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RafaelGP
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