Algebra equation

Question

19,297 is invested, part at 13% and the rest at 8%. if the interest earned from the amount invested at 13% exceeds the interest earned from the amount invested at 8% by 777.58, how much is the invested at each rate

Answer 1

0.13(19257 - x) = 0.08x + 777.58

Can you solve for x and answer?

Answer 2

Let x=$ invested at 13%

y = $ invested at 8%

.13x = interest earned on the amount invested at 13% (remember this amount is greater than the following)

.08y = interest earned on amount invested at 8%

x + y = 19297 (because the total invested is 19297)

.13x - .08y = 777.58 (the interest earned on 13% minus the interest earned at 8% differs by 777.58)

From here you have choices on how to solve the system.

If you do not like working with decimals multiply every term of the 2nd equation by 100 to clear the decimals:

13x - 8y = 77758

If you use the substitution method solve for either x or y and substitute into the other equation:

y = 19297 - x

So:

13x - 8(19297-x) = 77758

13x + 8x = 77758 + 8(19297)

21x = 232134

x = 232134/21 = $11054

And:

y = 19297 - 11054 = $8243

For confirmation and checking the result look here:

desmos.com/calculator/ofycegcrmp