If x is invested at 6%, then \(\displaystyle{8900}−{x}\) is invested at 10%.

The interest earned in one year is $674 so we can write:

\(\displaystyle{0.06}{x}+{0.10}{\left({8900}−{x}\right)}={674}\)

Solve for x:

\(\displaystyle{0.06}{x}+{890}−{0.10}{x}={674}\)

\(\displaystyle−{0.04}{x}+{89}={674}\)

\(\displaystyle−{0.04}{x}=−{216}\)

\(\displaystyle{x}={5400}\)

So, \(\displaystyle{8900}−{5400}=\${3500}\) was invested at 10%.

The interest earned in one year is $674 so we can write:

\(\displaystyle{0.06}{x}+{0.10}{\left({8900}−{x}\right)}={674}\)

Solve for x:

\(\displaystyle{0.06}{x}+{890}−{0.10}{x}={674}\)

\(\displaystyle−{0.04}{x}+{89}={674}\)

\(\displaystyle−{0.04}{x}=−{216}\)

\(\displaystyle{x}={5400}\)

So, \(\displaystyle{8900}−{5400}=\${3500}\) was invested at 10%.