[b][color=#bf9000]A SOLUTION to a linear inequality of 2 variables is an ordered pair ([i]x[/i], [i]y[/i]) whose coordinates satisfy the original inequality (i.e. make the original inequality true.) [/color][/b][br][br][color=#bf9000][b]For example, note the point (5, 3) shown in the applet below. [/b][/color][br][color=#1e84cc][b]The graph of the inequality [/b][/color][math]1x+1y\ge0[/math][color=#1e84cc][b] is also shown. [/b][/color][br][br][color=#bf9000][b](5, 3) is a SOLUTION [/b][/color][color=#1e84cc][b]to the inequality [/b][/color][math]1x+1y\ge0[/math] because [br][math]1\left(5\right)+1\left(3\right)\ge0[/math][br][math]5+3\ge0[/math][br][math]8\ge0[/math] ==> which is TRUE! [br][br]Any point ([i]x[/i], [i]y[/i]) that does NOT satisfy the original inequality (that is, does NOT make the original inequality true) is said NOT to be a solution to the given inequality. [br][br]How many solutions ([i]x[/i], [i]y[/i]) does the graph of an linear inequality of two variables have? [br]
Change the parameters [i]A[/i], [i]B[/i], and [i]C[/i] by using the sliders or by inputting values in the input boxes. Drag the [color=#bf9000][b]yellow point[/b][/color] around in the graph of this linear inequality. Algebraically show that this point is a SOLUTION to this linear inequality.
Pick a point ([i]x[/i], [i]y[/i]) that does NOT lie in the graph of the [color=#1e84cc][b]shaded blue region[/b][/color]. Show that this point is NOT a solution to this linear inequality.