Answer
$9\;lb$.
Work Step by Step
Total weight $T=17.4\;lb$.
Empty cooler weight $E=3.6\;lb$.
Ice weight $I=5\;lb$.
Let the fish weight $=F$
We can write.
$T=E+I+F$
Substitute all the values.
$17.4\;lb=3.6\;lb+5\;lb+F$
Simplify.
$17.4\;lb=8.6\;lb+F$
Subtract $8.6\;lb$ from both sides.
$17.4\;lb-8.6\;lb=8.6\;lb+F-8.6\;lb$
Simplify.
$8.8\;lb=F$
$9\;lb=F$.
Hence, the weight of the fish is about $9\;lb$.
