If x is invested at 5%, then 4000−x is invested at 4.5%.

The simple interest earned in 3 years is $547.50 so each year, the simple interest earned is $547.50/3=182.50 so we can write:

\(\displaystyle{0.05}{x}+{0.045}{\left({4000}−{x}\right)}={182.50}\)

Solve for x:

\(\displaystyle{0.05}{x}+{180}−{0.045}{x}={182.50}\)

\(\displaystyle{0.005}{x}={2.50}\)

\(\displaystyle{x}=\frac{{2.50}}{{0.005}}\)

\(\displaystyle{x}={500}\)

So, the Eric loaned $500 at 5% and $3,500 at 4.5%.

The simple interest earned in 3 years is $547.50 so each year, the simple interest earned is $547.50/3=182.50 so we can write:

\(\displaystyle{0.05}{x}+{0.045}{\left({4000}−{x}\right)}={182.50}\)

Solve for x:

\(\displaystyle{0.05}{x}+{180}−{0.045}{x}={182.50}\)

\(\displaystyle{0.005}{x}={2.50}\)

\(\displaystyle{x}=\frac{{2.50}}{{0.005}}\)

\(\displaystyle{x}={500}\)

So, the Eric loaned $500 at 5% and $3,500 at 4.5%.