Question

​$9500 is​ invested, part of it at 12​% and part of it at 9​%. For a certain​ year, the total yield is ​$1032.00. How much was invested at each​ rate?

Answer 1

Let's say the quantities are "a" and "b"
so... whatever "a" and "b" are, they add up to 9,500, thus

a + b = 9,500

now, "a" was invested at 12%, or 12/100
it yielded some amount, actually 12% of a, or 12/100 * a or 0.12a

the "b" amount was invested at 9%, or 9/100
it yielded 9% of b, or 9/100 * b or 0.09b

whatever 0.12a is, and 0.09b is, their sum or total yield was 1032
thus
0.12a + 0.09b = 1032

thus     $\bf \begin{cases} a+b=9,500\implies \boxed{b}=9,500-a\\\\ 0.12a+0.09b=1032\\ --------------\\ 0.12a+0.09(\boxed{9,500-a})=1032 \end{cases}$

solve for "a" to see how much was invested at 12%

what about "b"? well, b = 9500 - a