Question

$14,929 is invested, part at 9% and the rest at 6%. If the interest earned from the amount invested at 9% exceeds the interest earned from the amount invested at 6% by $1139.91, how much is invested at each rate? Round to two decimal places if necessary

Answer 1

Let x be the amount invested at 9% and y the amount invested at 6%.

Then we have:

I) x + y = $14,929

II) 0.09x - 0.06y = $1,139.91

First, we multiply equation II by 100:

9x - 6y = $113,991

Now, we add to it 6 times equation I:

15x = $203,565

x = $13,571

Now, we replace x in equation I to find y:

$13,571 + y = 14,929

y = $1,358