A. y > x + 4.....slope = 1....y int = (0,4)....x int = (-4,0)....it will be a dashed line...it will be shaded above the line...to graph, start at (-4,0)...and since slope is 1, go up 1 and to the right one....plot that point...then up 1 and to the right 1...plot that....keep doing this and u will cross the y axis at (0,4)....and just connect ur plotted points and u have ur line y > = x + 2....slope = 1...y int = (0,2)...x int = (-2,0)...it will be a solid line...it will be shaded above the line......to graph, start at (-2,0)...and since the slope is 1, go up 1 and to the right 1...plot that...and up 1 and to the right 1...plot that...u should cross the y axis at (0,2)....then just connect ur plotted points and u have ur line =============== B.To verify that the points D (-4,2) and E (-1,5) are solutions, sub in ur points into each of the inequalities. For them to be solutions, they have to work in both inequalities, not just one
subbing in (-4,2) y > x + 4 y > = x + 2 2 > -4 + 4 2 > = -4 + 2 2 > 0 (correct) 2 > = 2 (correct)
subbing in (-1,5) y > x + 4 y > = x + 2 5 > -1 + 4 5 > = -1 + 2 5 > 3 (correct) 5 > = 1 (correct) ====================== C. y > 3x - 4.....slope is 3....y int (where the line crosses the y axis) is -4.....x int (where the line crosses the x axis is 4/3 (or 1 1/3)...shading will be above the line....and it will be a dashed line. Once graphed, the farms that can raise these chickens are : A,E,D,F
