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Inequalities can be represented very well by constructing graphs to illustrate their solutions. EXAMPLE: y < 3 The graph for this is similar on one level to the graph for y = 3. y=3 y<3 The graphs for y > 3 and y ≥ 3 are similar to each other. Notice that the difference between these 2 graphs is the boundary line that separates the shaded portion from the unshaded portion. The first boundary does not include y = 3 but the second boundary does. y>3 y≥3 The following 5 graphs are similar however they all define a very distinct set of ordered pairs. y = 3 y ≤ 3 When the line is solid, the points on the line are a part of the solution. (-2,3) (5,3) (8,3) When the line is broken, the points on the line are x x (7,1) (-1,0) NOT a part of the solution. (5,-4) y y y > 3 y < 3 y ≥ 3 (8,1) (-2,9) (8,9) (7,5) (6,3) x x x (0,0) (5,-2) y y y Notice that when the equation changes to ‘x =‘ instead of ‘y =‘, the line becomes vertical instead of horizontal. This change is reflected when dealing with inequalities (<, >, ≤ or ≥) as well. x=3 x<3 When both variables (x and y) are used, the line will be neither horizontal nor vertical. It will be oblique (inclined). That can mean a positive (inclined up) or negative (inclined down) slope. When there is an equation with both variables, it is necessary to make a table of values to find 3 ordered pairs. These ordered pairs will establish the direction of the oblique line. STEP 1: Isolate y x+y=5 Example: x + y = 5 y = -x + 5 STEP 2: Construct the Table of Values. x y = -x + 5 y (x,y) 0 y = -(0) + 5 5 (0,5) y= 5 2 y = -(2) + 5 3 (2,3) y= 3 -2 y = -(-2) + 5 7 (-2,7) y= 7 x+y=5 x+y>5 If the ‘ = ‘ is changed to a ‘ < ‘, the solid line will become a broken line. At that point it is necessary to decide on shading one of the 2 sides of the line. To do this we can choose a point from one side and test it in the inequality. Let’s test (0,0) in the inequality. x + y > 5 (0,0) 0+0>5 0 > 5 False If (0,0) is causes the inequality to be false then so will all of the points on that side of the line. We can test other points to prove this. Test (2,1) Test (-5,6) x+y>5 x+y>5 2+1>5 (-5) + 6 > 5 3 > 5 False 1 > 5 False Test (-3,-5) x+y>5 (-3) + (-5) > 5 The point is that whenever a point on this -8 > 5 False side of the line is tested, the result will be false. This means that these are not ordered pairs that make the statement, ‘x + y > 5’, true. This leads us to the obvious question: Which ordered pairs can make this inequality TRUE. If we choose ordered pairs from the other side of the line, we may get a different result. Let’s choose (5,5). If (5,5) makes the inequality TRUE, so should the other points on the same side of the boundary. We can prove that this is the case for (0,8), (6,0) and (10,-3). We can express the solution for the inequality by shading the side with the true ordered pairs. Notice that the boundary line is broken. This indicates that ordered pairs that fall on the line will make the inequality false. In other words, those ordered pairs are not a part of the solution. Test (5,5) Test (0,8) Test (6,0) Test (10,-3) x+y>5 x+y>5 x+y>5 x+y>5 5+5>5 0+8>5 6+0>5 10 + (-3) > 5 10 > 5 True 8 > 5 True 6 > 5 True 7 > 5 True When deciding on which side of the line to shade, it is NOT really necessary to test a bunch of ordered pairs. You need only test one. Choose an ordered pair that is clearly NOT on the boundary. Test to see if it makes the inequality TRUE or FALSE. If the ordered pair makes the If the ordered pair makes the inequality TRUE, shade all of the inequality FALSE, shade all of points on that side. the points on the OTHER side. This way no matter whether you choose a point from true side or not, the true side will end up being shaded. (Top side in this case) With everything that was discussed so far, it is necessary to demonstrate the steps necessary to construct a linear inequality graph. Example 1: 4x + 3y < 12 To graph an inequality, we must determine the boundary We can find his by changing the inequality to an equation. 4x + 3y < 12 becomes 4x + 3y = 12 to find the boundary. If there are no fractions in the inequality, we can isolate y right away. 3y = -4x + 12 3y 4x 12 3 3 3 4x y 4 3 Now it is time to construct a table of values to find 3 ordered pairs. Notice that the denominator of the x co-efficient is 3. This means that it would be very convenient to choose x-values that are multiples of 3 … like –6, -3, 0, 3 , 6, 9, etc. x y 4x 4 y With the 3 points located on the 3 graph, we can see the direction of the 4(0) line. We must decide whether the line 0 y 4 4 should be a solid line ( ) or a broken 3 044 line ( ). To make this decision, it is 3 4(3) 0 necessary to observe the inequality y 4 3 sign in the original inequality. 4 4 0 6 4(6) -4 y 4 3 8 4 4 < or > requires a broken line ≤ or ≥ requires a solid line 4x + 3y < 12 This example has a ‘less than’ sign which means the boundary will be a broken line 4x + 3y < 12 Test (0,0) in 4x + 3y < 12 Where must we shade the 4(0) + 3(0) < 12 graph to indicate the points 0 + 0 < 12 that will make the equation true? To find out we must 0 < 12 TRUE choose and test a non- boundary ordered pair in the inequality. (0,0) makes the inequality true. All of the points on this side would do the same. This is the side that must be shaded. Example 2: 3x - 5y ≥ 7 Find boundary of 3x – 5y = 7 5 y 3x 7 5 y 3x 7 ≤ or ≥ requires a solid Isolate y line so the boundary 5 5 5 line should be solid in 3x 7 this case. y 5 x y 3x 7 y 5 3( 0) 7 7 0 y -1.4 5 5 1.4 3( 4) 7 12 7 4 y 1 5 5 1 3( 1) 7 3 7 -1 y -2 5 5 2 Test (0,0) in 3x - 5y ≥ 7 (0,0) makes the inequality 3(0) + 5(0) ≥ 7 false. All of the points on this 0+0≥7 side would do the same. The points on the other side must 0 ≥ 7 FALSE make it true. It is therefore, the other side that must be shaded. Solve the System of Inequalities: -x + 3y ≤ 3 4y + 3x + 9 > 0 Change the inequality signs to equal signs to find the boundary for each. -x + 3y = 3 4y + 3x + 9 = 0 Isolate y for each. 3y x 3 4 y 3x 9 3y x 3 4 y 3x 9 3 3 3 4 4 4 1 3x 9 y x1 y 3 4 1 3x 9 y x1 y 3 4 x y 1 x1 y x y 3x 9 y 3 4 1 3( 0) 9 9 0 y ( 0) 1 1 0 y -2.25 3 4 4 0 1 1 2.25 1 3(1) 9 3 9 -3 y ( 3) 1 0 1 y -3 3 4 4 1 1 0 3 1 3( 3) 9 9 9 6 y (6) 1 3 -3 y 0 3 4 4 21 3 0 (0,1) (0,-2.25) (-3,0) (1,-3) (6,3) (-3,0) -x + 3y ≤ 3 4y + 3x + 9 > 0 Test (0,0) in -x + 3y ≤ 3 Test (0,0) in 4y + 3x + 9 > 0 (0,1) (0,-2.25) -(0) + 3(0) ≤ 3 4(0) + 3(0) + 9 > 0 (-3,0) (1,-3) 0+0≤3 0+0+9>3 (6,3) (-3,0) 0 ≤ 3 TRUE 9 > 0 TRUE The solution to the system of inequalities is the area on the grid that contains ordered pairs that makes both inequalities true. Of the four regions defined by the intersecting boundary lines, only one of those regions will be true for BOTH inequalities. Some of the ordered pairs in this region includes: (0,0), (10,2), (5,-4), (4,1) and (-1,-1)

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