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mixas84 [53]
3 years ago
15

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?
Mathematics
1 answer:
Blababa [14]3 years ago
3 0

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:

A: We'll start with the equation for the total charge. Juan is saying he will pay a one time fee of $30, meaning there is 30 dollars added to what ever the hourly wage is. This can be represented by the +30. Now if "h" represents the variable for which the hourly wage will be calculated, and he pays $45 dollars per hour this will be represented as 45h. <em>As an example if I pay you 2 dollars per hour using the same variable "h", this would be represented as 2h. So if you worked for two hours you would get 4 dollars, this is proven by the fact the 2(2) (remeber im replacing the "h" with the hours you worked) obviously 2 times 2 is 4 proving my point. </em>This information will give you the equation you see above.

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This results in the answer of:

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