$12312 is invested, part at 14% and the rest at 10%. if the interest earned from the amount invested at 14% exceeds the interest earned from the amount invested at 10% by $512.16, how much is invested

Question

$12312 is invested, part at 14% and the rest at 10%. if the interest earned from the amount invested at 14% exceeds the interest earned from the amount invested at 10% by $512.16, how much is invested at each rate? 

Answer 1

Two variables can be used if you want:

Let x = amount invested at 14%

y = amount invested at 10%

x + y = 12312 (total amount invested is 12312)

.14x represents the interest earned on that amount

.10y represents the interest earned on that amount

.14x - .10y = 512.16

Solving the 1st equation for y in order to use the substitution method:

y= 12312 - x

.14x - .10(12312 - x) = 512.16

At this point multiply every term by 100 to clear the decimals:

14x - 10 (12312 - x ) =51216

24x = 51216+123120

x = 174336/24 = 7264 (the amount invested at 14%)

12312-7264=5048 (amt at 10%)

Check:

.14(7264)=1016.96

.10(5048)=504.80

Difference = 512.16