Question 700985
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Yes for all.


a.  You need a pair of single variable inequalities in the same variable such that the intersection of the solution set is a single value.  If you are restricted to two-variable inequalities, then the answer is No, the solution set cannot consist of a single point. *[tex \LARGE x\ \leq\ 1] and *[tex \LARGE x\ \geq\ 1] for example. 


b.  You need a pair of two-variable linear inequalities such that both boundary lines are included in the solution set of each and the intersection of the two solution sets is the boundry line.   *[tex \LARGE y\ \leq\ x] and *[tex \LARGE y\ \geq\ x] for example.


d.  (purposely out of order).  You need a pair of two-variable linear inequalities such that the boundary lines are parallel lines and the intersection of the two solution sets is the empty set.  *[tex \LARGE y\ \geq\ x\ +\ 1] and *[tex \LARGE y\ \leq\ x\ -\ 1] for example.


c.  Any pair of two-variable linear inequalities that do not fit the criteria of b and d above.   *[tex \LARGE x\ -\ y\ >\ 3]  and *[tex \LARGE x\ +\ y\ <\ 5] for example.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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