Question

6300 is invested, part of it at 10% and part of it at 9%. For a certain year, the total yield is $598.00. How much was invested at 10%? At 9%?

Answer 1

let's say "a" amount was invested at 10%
and "b" amount was invested at 9%

whatever "a" and "b" are, 6,300 = a + b

now, the 10% one, yielded 10/100 * a, or 0.1a
the 9% one, yielded 9/100 * b or 0.09b

both those yields added together gave 598.00
so
`\bf \begin{cases} 6,300=a+b\to \boxed{6,300-a}=b \\\\ 0.1a+0.09b=598\\ --------------\\ 0.1a+0.09\boxed{b}=598 \end{cases}`

do the substitution and solve for "a", to see how much the 10% was

what about "b"? well, 6,300 - a = b