<p>Linear Systems – Graphing Inequalities Worksheet Answers Graph each system and find the solution area</p><p> x  y  0 dashed line 1. 2x  y  4 dashed line</p><p>I solved for “y” and used slope - B intercept to graph the lines…</p><p> x  y  0 1 m   y  x  0 1 A y  x  0 b  0,0</p><p>2x  y  4  2 m   2x  2x 1 y  2x  4 b  0,4</p><p>TESTS POINTS :</p><p>A ( 3 , 0 ) B ( 0 , 6 ) 20 6  4 3  0  0 0  6  0 FALSE TRUE 0  6  4 TRUE 3  0  6  0 6  4</p><p>** Since test point B has two trues, that area is the solution set… 3x  y  2 solid 2. x  4y  5 dashed</p><p>I solved for “y” and used slope − intercept to graph</p><p>3x  y  2 3  3x  3x m  1  y  3x  2 B A b  0,2 y  3x  2</p><p> x  4y  5  x  x 1 m  4y  x  5 4 1 5 b  0 , 1.25 y  x  4 4</p><p>TEST POINTS :</p><p>A ( 0 , 0 ) B ( − 3 , 0 )</p><p>30 0  2 3 3 0  2  3  40  5 FALSE TRUE TRUE 0  2  9  2  3  5</p><p>** Since point B has two TRUE, that is your solution area… 2x  3y  4 dashed 3. 4x  y  7 solid</p><p>Again, solve for “y” and use C slope – intercept to graph</p><p>2x  3y  4  2x  2x 2 B m   3y  2x  4 3 A 2 4 b  0 , 1.33 y  x  3 3</p><p>4x  y  7  4 m   4x  4x 1 y  4x  7 b  0,7</p><p>TEST POINTS A ( 0 , 0 ) 20 30  4 40 0  7 TRUE FALSE 0  4 0  7</p><p>B ( 4 , 0 ) 24 30  4 8  0  4 FALSE 8  4</p><p>C ( 3 , 5 )</p><p>23 35  4 43 5  7 6 11  4 TRUE 12  5  7 TRUE  5  4 17  7</p><p>Point C has two TRUE so shade that area x  3y  6 solid 4. 2x  4y  12 solid</p><p> x  3y  6 1 3y  x  6 m  3 1 y  x  2 b  0, 2 3</p><p>2x  4y  12 1 A  4y  2x 12 m  2 1 y  x  3 b  0,3 2 B</p><p>TEST POINTS</p><p>A ( 0 , 0 ) 0  30  6 20 40  12 TRUE FALSE 0  6 0  12</p><p>B ( 0 , − 5 )</p><p>0  3 5  6 20 4 5  12 TRUE TRUE 15  6 20  12</p><p>Point B has 2 TRUE, shade that area…</p>