We invested $10000 into two bank accounts. One account earns 14% per year, the other account earns 8% per year. How much did we invest into each account if after the first year, the combined interest from the two accounts is $1238?

Answers

Answer 1

Answer:

$7,300 was invested in the 14% account while $2,700 was invested in the 8% interest account

Step-by-step explanation:

Let the amount invested in first account be $x while the amount for second account is $y

Mathematically;

x + y = 10,000 •••••••••(i)

Interest on first account

= 14% of x = 14/100 * x

Interest on second account

= 8% of y= 8/100 * y

Adding all together will give total

14/100 * x + 8/100 * y = 1238

Multiply through by 100

14x + 8y = 123800 •••••••(ii)

From equation 1, x = 10,000 - y

Insert this into second equation

14(10,000-y) + 8y = 123,800

140,000 - 14y + 8y = 123,800

140,000 - 6y = 123,800

6y = 140,000 - 123,800

6y = 16,200

y = 16,200/6

y = $2,700

Recall

x + y = 10,000

x = 10000 - y

x = 10,000 - 2,700

x = $7,300


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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Complete question :

The temperature at 5:00 a.m. was –6 °F. By 3:00 p.m., the temperature had risen to a high of 36 °F. If the temperature at 8:00 p.m. decreased from the high by the change in temperature in Fahrenheit degrees from 5:00 a.m. to 3:00 p.m., what was the temperature at 8:00 p.m.?

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Step-by-step explanation:

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Answers

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ОО

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Answers

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