Question

$19,670 is invested, part at 10 % and the rest at 6 %. If the interest earned from the amount invested at 10% exceeds the interest earned from the amount invested at 6 % by $681.08, how much is invested at each rate? (Round to two decimal places if necessary.)​

Answer 1

$1,762 is invested at 10%, and approximately $21,432 is invested at 6%.

Explanation:

amount invested at 10% is "x" and the amount invested at 6% is "y."

given pieces of information:

1. The total amount invested is $19,670: x + y = 19,670.

2. The interest earned from the amount invested at 10% exceeds the interest earned from the amount invested at 6% by $681.08: 0.10x - 0.06y = 681.08.

Now, we can solve this system of linear equations. You can use the substitution or elimination method. im using the elimination method:

First, multiply the second equation by 25 to make the coefficients of "y" in both equations equal:

2.5x - 1.5y = 17,027.

subtract the second equation from the first equation:

(1) x + y = 19,670

(2) 2.5x - 1.5y = 17,027

Subtract (1) from (2):

(2.5x - x) - (1.5y - y) = 17,027 - 19,670

1.5x = -2,643

Now, divide both sides by 1.5 to solve for x:

x = -2,643 / 1.5

x = -1,762

Now that we have the value of x, we can find y using the first equation:

x + y = 19,670

-1,762 + y = 19,670

Add 1,762 to both sides:

y = 19,670 + 1,762

y = 21,432

So, approximately $1,762 is invested at 10%, and approximately $21,432 is invested at 6%.