Answer
$160$ pounds of haddock, $340$ pounds of sea bass, and $60$ pounds of red snapper.
Work Step by Step
Step 1. Assume he caught $x$ pounds of haddock, $y$ pounds of sea bass, and $z$ pounds of red snapper.
Step 2. Based on the conditions given, we can set up the following system of equations:
$\begin{cases} x+y+z=560\hspace3cm(total-weight) \\1.25x+0.75y+2.00z=575\hspace1cm(total-value) \\1.25x+2.00z=320\hspace2cm(haddock+snapper) \end{cases}$
Step 3. Simplify the equations by removing the decimals:
$\begin{cases} x+y+z=560 \\125x+75y+200z=57500 \\125x+200z=32000 \end{cases}$
Step 4. Further simplify the equations by cancelling common factors
$\begin{cases} x+y+z=560 \\5x+3y+8z=2300 \\5x+8z=1280 \end{cases}$
Step 5. The last equation gives $5x=1280-8z$, use it for substitution into the first and second equations:
$\begin{cases} 1280-8z+5y+5z=2800 \\1280-8z+3y+8z=2300 \\5x+8z=1280 \end{cases}$
or
$\begin{cases} 5y-3z=1520 \\3y=1020 \\5x+8z=1280 \end{cases}$
Step 6. The middle equation gives $y=340$, thus we can get $z=60$ and $x=160$
Step 7. Conclusion: he caught $160$ pounds of haddock, $340$ pounds of sea bass, and $60$ pounds of red snapper.
