SOLUTION: $14,700 is invested, part at 3% and the rest at 2%. If the interest earned from the amount invested at 3% exceeds the interest earned from the amount invested at 2% by $359.00, how

Question 1156306: $14,700 is invested, part at 3% and the rest at 2%. If the interest earned from the amount invested at 3% exceeds the interest earned from the amount invested at 2% by $359.00, how much is invested at each rate? (Round to two decimal places if necessary.)
Define variables x and y and set up a system of two linear equations that represents the information given in the problem.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

x = amount (in dollars) invested at 3%
y = amount (in dollars) invested at 2%
There is $14700 invested total, so x+y = 14700 which solves to y = 14700-x. We'll use this equation later in a substitution step.

Through the simple interest formula, i = P*r*t, we can say
i = P*r*t = x*0.03*1 = 0.03x = amount of annual interest from the 3% account

and also
i = P*r*t = y*0.02*1 = 0.02y = amount of annual interest from the 2% account

We're told that the amount of interest from the 3% account exceeds the other account's interest by $359, so,
0.03x = 0.02y+359

0.03x = 0.02y+359
0.03x = 0.02(14700-x)+359 .... Substitution; plug in y = 14700-x
0.03x = 0.02(14700)+0.02(-x)+359 ... distribute
0.03x = 294-0.02x+359
0.03x = -0.02x+653
0.03x + 0.02x = 653 ... add 0.02x to both sides
0.05x = 653
x = 653/0.05 ..... divide both sides by 0.05
x = 13060

Once you've figured out the value of x, you can find the value of y
y = 14700 - x
y = 14700 - 13060
y = 1640

Answers:
$13,060 invested at 3%
$1,640 invested at 2%

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