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$10,637 is invested, part at

13% and the rest at

6%

.

If the interest earned from the amount invested at

13% exceeds the interest earned from the amount invested at

6% by

$880.07, how much is invested at each rate? (Round to two decimal places if necessary.)

Verified Answer :

now... we know that, whatever the earned interest was,
the quantity of "a" is bigger than "b", by 880.07 bucks,
or
a = b + 880.07

whatever "b" is, add 880.07 to be able to level with "a"
so \(\bf \begin{cases} a+b=10,637\to a=\boxed{10,637-b} \\\\ \frac{13}{100}a=\frac{6}{100}b+880.07\\ or\\ 0.13a = 0.06b + 880.07\\ --------------\\ 0.13(\boxed{10,637-b}) = 0.06b + 880.07 \end{cases}\)

solve for "b", to see how much that was at 6%,

what about "a"? well, a = 10,637 - b

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