Question

Piotr Galkowski invested some money at 3.5% simple interest, and $5000 more than three times this amount at 4%. He earned $1440 in annual interest. How much did he invest at each rate?

Answer 1

so... Piotr invested two amounts, say "a" and "b", at 3.5% and 4% respectively

whatever 3.5% of a is, and whatever 4% of b is, it ended up as 1440

now, we know that "b" amount is "$5000 more than three times" than "a" amount

so three times "a" is 3*a or 3a, now, 5000 more than that is 3a + 5000

now, assuming this is for a year alone,
how much is 3.5% of "a", well, 3/100 * a, or 0.035a
how much is 4% of "b", well, 4/100 * a, or 0.04b

so.. whatever those amounts yielded are, they ended up as 1440
so
`\bf \begin{cases} \textit{3.5\% of a}\implies (3.5)/(100)\cdot a\implies 0.035a \\\\ \textit{4\% of b}\implies (4)/(100)\cdot b\implies 0.04b\\ --------------\\ 0.035a+0.04\boxed{b}=1440 \\\\ however\qquad b=\boxed{3a+5000} \end{cases}`

do the substitution, and solve for "a", to see how much he invested at 3.5%

how much is "b"? well, b = 3a + 5000