Let x be the amount invested at 10%, yy be the amount invested at 7%, and z be the amount invested at 8%.

The total investment is $200000:

\(\displaystyle{x}+{y}+{z}={200000}{\left({1}\right)}\)

The annual income is $16,000:

\(\displaystyle{0.10}{x}+{0.07}{y}+{0.08}{z}={16000}{\left({2}\right)}\)

The combined investment in alternatives 2 and 3 should be triple the amount invested in alternative 1 so:

\(\displaystyle{y}+{z}={3}{x}\)

or

\(\displaystyle−{3}{x}+{y}+{z}={0}−{3}{x}+{y}+{z}={0}{t}{a}{g}{\left\lbrace{3}\right\rbrace}\$\$\)

Subtract each side of (1) and (3) then solve for xx:

\(\displaystyle{4}{x}={200000}\)

\(\displaystyle{x}={50000}\)

Substitute x=50000 to (2) and simplify to obtain (4):

\(\displaystyle{0.10}{\left({50000}\right)}+{0.07}{y}+{0.08}{z}={16000}\)

\(\displaystyle{5000}+{0.07}{y}+{0.08}{z}={16000}\)

\(\displaystyle{0.07}{y}+{0.08}{z}={11000}{\left({4}\right)}\)

Substitute x=50000 to (1) and simplify to obtain (5):

\(\displaystyle{50000}+{y}+{z}={200000}\)

\(\displaystyle{y}+{z}={150000}{\left({5}\right)}\)

Eliminate zz. Multiply (5) by 0.08 to obtain (6):

\(\displaystyle{0.08}{y}+{0.08}{z}={12000}{\left({6}\right)}\)

Subtract each side of (4) and (5) then solve for x:

\(\displaystyle−{0.01}{y}=−{1000}\)

\(\displaystyle{y}={100000}\)

Solve for zz using (5):

\(\displaystyle{100000}+{z}={150000}\)

\(\displaystyle{z}={50000}\)

So, $50000 was invested at 10%, $100000 was invested at 7%, and $50000 was invested at 8%.

The total investment is $200000:

\(\displaystyle{x}+{y}+{z}={200000}{\left({1}\right)}\)

The annual income is $16,000:

\(\displaystyle{0.10}{x}+{0.07}{y}+{0.08}{z}={16000}{\left({2}\right)}\)

The combined investment in alternatives 2 and 3 should be triple the amount invested in alternative 1 so:

\(\displaystyle{y}+{z}={3}{x}\)

or

\(\displaystyle−{3}{x}+{y}+{z}={0}−{3}{x}+{y}+{z}={0}{t}{a}{g}{\left\lbrace{3}\right\rbrace}\$\$\)

Subtract each side of (1) and (3) then solve for xx:

\(\displaystyle{4}{x}={200000}\)

\(\displaystyle{x}={50000}\)

Substitute x=50000 to (2) and simplify to obtain (4):

\(\displaystyle{0.10}{\left({50000}\right)}+{0.07}{y}+{0.08}{z}={16000}\)

\(\displaystyle{5000}+{0.07}{y}+{0.08}{z}={16000}\)

\(\displaystyle{0.07}{y}+{0.08}{z}={11000}{\left({4}\right)}\)

Substitute x=50000 to (1) and simplify to obtain (5):

\(\displaystyle{50000}+{y}+{z}={200000}\)

\(\displaystyle{y}+{z}={150000}{\left({5}\right)}\)

Eliminate zz. Multiply (5) by 0.08 to obtain (6):

\(\displaystyle{0.08}{y}+{0.08}{z}={12000}{\left({6}\right)}\)

Subtract each side of (4) and (5) then solve for x:

\(\displaystyle−{0.01}{y}=−{1000}\)

\(\displaystyle{y}={100000}\)

Solve for zz using (5):

\(\displaystyle{100000}+{z}={150000}\)

\(\displaystyle{z}={50000}\)

So, $50000 was invested at 10%, $100000 was invested at 7%, and $50000 was invested at 8%.