Since the test point (0, 0) does not satisfy the inequality we will shade the region that does not contain the point (0, 0). All points on the line and in the shaded region satisfy the inequality 3x 4y 8. x y 5 Test point (0, 0) 5 5 3x 4y 8 5 EXAMPLE 2 Graph x 2y 4. Solution 1) Since the inequality symbol is , graph a dotted boundary line, x 2y 4. (This means that the points on the line are not part of the solution set.) 2) Choose a test point not on the line and substitute it into the inequality to determine whether it makes the inequality true. If the inequality does not include the equals condition, the boundary line is dotted. Points on this line make the inequality false. x y 5 t t 5 5 5 Test Point Substitute into x 2y 4 (0, 0) (0) 2(0) 4 0 4 True Since the test point (0, 0) satisfi es the inequality, shade the region containing that point. All points in the shaded region satisfy the inequality x 2y 4. x y 5 x 2y 4 Test point (0, 0) 5 5 5 YOU TRY 1 Graph each inequality. a) 2x y 4 b) x 4y 12 If we write the inequality in slope-intercept form, we can decide which region to shade without using test points. SECTION www.mhhe.com/messersmith 4.4 Linear and Compound Linear Inequalities in Two Variables 189
messersmith_power_intermediate_algebra_1e_ch4_7_10
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