Question

you invested $7000 between two accounts paying 3% and 9% annual interest, respectively. if the total interest earned for the year was $390, how much was the invested at each rate?

Answer 1

a = amount invested at 3%

how much is 3% of "a"? (3/100) * "a", namely 0.03a.

b = amount invested at 9%

how much is 9% of "b"? (9/100) * "b", namely 0.09b.

we know the total amount invested is 7000, so whatever "a" and "b" might be, we know that a + b = 7000.

we also know that the yielded amount in interest is 390, so if we simply add their interest, that'd be 0.03a + 0.09b.

`a+b=7000\implies b=7000-a \\\\[-0.35em] ~\dotfill\\\\ \stackrel{ \textit{sum of their interest} }{0.03a+0.09b}~~ = ~~390\implies \stackrel{\textit{substituting from above}}{0.03a+0.09(7000-a)}=390 \\\\\\ 0.03a+630-0.09a=390\implies -0.06a+630=390\implies 630=390+0.06a \\\\\\ 240=0.06a\implies \cfrac{240}{0.06}=a\implies \boxed{4000=a}\hspace{5em}\stackrel{ 7000~~ - ~~4000 }{\boxed{b=3000}}`