How do you graph the system of linear inequalities #2x+y<6# and #y> -2#?

#2x+y<6# and #y > -2#
First consider the related linear equations:
#2x+y=6 rarr y = -2x+6#
and
#y = -2#

#y = -2x+6#
is a line with y-intercept at #(0,6)# and a slope of #(-2)# (which means that for every unit increase in #x#, #y# decreases by #2#)

#y=-2#
is a horizontal line through the point #(0,-2)#

It is fairly simple to plot these lines;
then we just need to determine which sides of these lines need to be combined for the given inequalities
Since
#y>-2#
the points of the Range must be above the horizontal line

Since
#2x+y<6#
the points of the Range must be below the line for #2x+y=6# (since if #y_1# is in the Range and #y_2<y_1# then #y_2# must also be in the Range).

To graph the system of linear inequalities \(2x + y < 6\) and \(y > -2\), follow these steps: 1. Graph the line \(2x + y = 6\). To do this, rearrange the equation in slope-intercept form (\(y = mx + b\)): \[y = -2x + 6.\] Plot the y-intercept at 6 and use the slope of -2 to find another point. Draw a dashed line because the inequality is \(2x + y < 6\) (not \(\leq\)). 2. Graph the line \(y = -2\). This is a horizontal line passing through -2 on the y-axis. Since the inequality is \(y > -2\), the line should be drawn as a dashed line since it does not include the points on the line. 3. Shade the region that satisfies both inequalities. This is the area below the line \(2x + y = 6\) (since \(2x + y < 6\)) and above the line \(y = -2\) (since \(y > -2\)). The shaded region represents the solution to the system of inequalities.