Aboy drags a suitcase along the ground with a force of 100 n. if the frictional force opposing the motion of the suitcase is 50 n, what is the resultant forward force on the suitcase?

Fortunately, 'force' is a vector.  So if you know the strength and direction
of each force, you can easily addum up and find the 'resultant' (net) force.

When we talk in vectors, one newton forward is the negative of
one newton backward.   Hold that thought, while I slog through
the complete solution of the problem.

            (100 N forward) plus (50 N backward)

        =  (100 N forward) minus (50 N forward)

        =           50 N forward .

That's it.
Is there any part of the solution that's not clear ?

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