By A. S. Solodovnikov
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Evidently, the point A of the region ,R" is a vertex if and only if if belongs to two dijjerent boundary lines. l' the oriqinal system (1). R" cannot be greater than that either. So the number q{ the rertices is finite. Remark. It follows from the above that. _- the system has no solutions (it is incompatible), Example 1. \ On solving the subsystems + y + I == O} x - 2y - 2 == 0 x 36 x + y + 1== O} 2x - y - 4 == 0 2 O} ,x - 2y - = 2x- y-4==O l (all of them prove regular) we find three points: (O~ - 1), (1, - 2).
11'. v = 0 in the system (1). 41 Then we have the system of inequalities bty b2 y + Cl + C2 ~ 0 bmy + c; ~ 0 ~ 0 (12) in one unknown y which is easy to solve *. ff is also empty) or a point, a segment or a ray (but it cannot be the whole of the y-axis, for otherwise ff wouldbe the whole of the plane, which is impossible). I! is not parallel to the y-axis). Example 5. :ff of the system x+ Y-l~O} -x2x y+2~O + 2y + 3 ~ 0 I t is- easily seen that the system is not normal and that 9) straight line IS the x+y==O (not parallel to the j-axis).
Which is contrary to the definition of a vertex. This compels us to make some obvious changes in the method of finding vertices described in Subsection 2' of Section 5. \' + Ct Z + lit = amx + hmr + CmZ + dm = O} (4) 0 45 rather than two, provided the solution (x, y, z) of the subsystem is unique. f the system (4) and pick out those which satisf» the original system (1). The theorem of Subsection 2° of Section 5 also remains valid; the changes to be made in the proof of the theorem are obvious.