Question 245427
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First graph the line *[tex \LARGE y\ =\ 2x\ -\ 5] but, since the original inequality is *[tex \LARGE <] rather than *[tex \LARGE \leq], draw a dashed line instead of a solid line.


Next, select any point NOT on the line.  This line does not pass through the origin, so *[tex \Large \left(0,0\right)] is an outstanding choice.  Substitute the coordinate values from the point you selected into the original inequality.


If the result is a true statement, then shade the side of the line that contains the point selected.


If the result is a false statement, shade the other side of the line.


Your other inequality is not a straight line, but that doesn't change the procedure much.  In this case, when you change it to an equation in order to graph the boundary, you should recognize that this fits the following pattern:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - h)^2 + (y - k)^2 = r^2]


which is the equation of a circle with center at *[tex \Large \left(h,k\right)] and radius *[tex \Large r]


So, draw your circle, again using a dashed line because the original inequality is *[tex \LARGE <] rather than *[tex \LARGE \leq].


Again, select a test point.  Since the circle does not pass through the origin, *[tex \Large \left(0,0\right)] is again an outstanding choice.  In this case, you will find that *[tex \Large \left(0,0\right)] is inside the circle.  If the coordinates *[tex \Large \left(0,0\right)] make the original inequality true, shade the inside of the circle.  If the coordinates make the original inequality false, shade the outside area.


You should now have two regions shaded.  Where those two regions overlap is the desired solution set.  Note that the dashed lines indicate that points on the straight line or on the circle are NOT included in the solution set to your system.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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