\(-x + 10 \ \textless \ 0\)
Transposing the inequation gives:
\(-x \ \textless \ -10\)
Since we want to get rid of the negatives on both sides, we must flip the sign:
\(x \ \textgreater \ 10\)
If you transpose the equation to make RHS be 0:
\(x - 10 \ \textgreater \ 0\)
That seems to satisfy x - 10 (could be wrong though...)
But I'm not exactly clear on what you want as the options above don't seem to give that much context :I
