Respuesta :
y > 5x + 5 : slope = 5, y int = (0,5), x int = (-1,0)
since there is no equal sign, it a dashed line...and since it is greater then, it is shaded above the line.
y > -1/2x + 1 : slope = -1/2, y int = (0,1), x int = (2,0)
since there is no equal sign, it is a dashed line...and since it is greater then, it is shaded above the line.
is (-2,5) included in the solution ?
Lets find out.....this solution would have to satisfy BOTH inequalities for it to be a solution.
So we sub and check...
y > 5x + 5
5 > 5(-2) + 5
5 > -10 + 5
5 > -5...correct...now we check the other one
y > -1/2x + 1
5 > -1/2(-2) + 1
5 > 1 + 1
5 > 2....correct
and since it satisfies both inequalities, it IS a solution
since there is no equal sign, it a dashed line...and since it is greater then, it is shaded above the line.
y > -1/2x + 1 : slope = -1/2, y int = (0,1), x int = (2,0)
since there is no equal sign, it is a dashed line...and since it is greater then, it is shaded above the line.
is (-2,5) included in the solution ?
Lets find out.....this solution would have to satisfy BOTH inequalities for it to be a solution.
So we sub and check...
y > 5x + 5
5 > 5(-2) + 5
5 > -10 + 5
5 > -5...correct...now we check the other one
y > -1/2x + 1
5 > -1/2(-2) + 1
5 > 1 + 1
5 > 2....correct
and since it satisfies both inequalities, it IS a solution
y > 5x + 5 . . . (1)
y > -1/2x + 1 . . . (2)
Part A: The graph of the system are two straight dotted lines with the first line passing through points (-1, 0) and (0, 5) with the region above the line shaded and the second line passing through (0, 1) and (2, 0) with the region above the line shaded. The two lines intersect at point (-8/11, 15/11).
Part B: The point (-2, 5) is included in the solution area.
5 > 5(-2) + 5 and 5 > -1/2(2) + 1
5 > -10 + 5 and 5 > -1 + 1
5 > -5 and 5 > 0
y > -1/2x + 1 . . . (2)
Part A: The graph of the system are two straight dotted lines with the first line passing through points (-1, 0) and (0, 5) with the region above the line shaded and the second line passing through (0, 1) and (2, 0) with the region above the line shaded. The two lines intersect at point (-8/11, 15/11).
Part B: The point (-2, 5) is included in the solution area.
5 > 5(-2) + 5 and 5 > -1/2(2) + 1
5 > -10 + 5 and 5 > -1 + 1
5 > -5 and 5 > 0