$14,038 is invested, part at 14% and the rest at 12%. If the interest earned from the amount invested at 14% exceeds the interest earned from the amount invested at 12% by $984.34, how much is invested at each rate? (Round to two decimal places if necessary.)

Answers

Answer 1

Answer:

492.99

Step-by-step explanation: x = amount invested at 12%

y = amount invested at 9%

---

x + y = 20287

0.12x = 0.09y + 492.99

---

put the system of linear equations into standard form

---

x + y = 20287

0.12x - 0.09y = 492.99

Answer 2

The amount invested at 14% is $4144 and the amount invested at 12% is $9894.

What is simple interest?

Simple interest is a method of calculating the interest charge. Simple interest can be calculated as the product of principal amount, rate and time period.

Simple Interest = (Principal × Rate × Time) / 100

We are given that;

Investment at 14%= $14,038

Investment at 12%= $984.34

Now,

Let x be the amount invested at 14% and y be the amount invested at 12%. Then we have:

x + y = 14038 (total amount invested)

0.14x - 0.12y = 984.34 (difference in interest)

We can solve this system of equations by substitution or elimination. For example, by elimination, we can multiply the first equation by 0.12 and subtract it from the second equation to get:

0.02x = 82.88

x = 4144

Then we can plug x into the first equation and get:

y = 14038 - 4144

y = 9894

Therefore, the simple interest will be $4144 and $9894.

Learn more about simple interest here;

https://brainly.com/question/1548909

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Answers

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