Respuesta :

Answer:

  • a) (0,0), (-4, 1), (-1,2)
  • b) None

Step-by-step explanation:

  • There are two ways of solving similar problems. One- substitute x and y with respective coordinates and verify if the inequalities are correct. Two- graph the lines and see if the points are in the intersection zones.
  • We'll go with the first method.
  • Remember both inequalities should be correct for the same point to accept it as a solution.

Given points

  • (0,0), (-4,1), (-1,2)

a) The system of inequalities

  • y > x - 1
  • y < -1/2x + 2

Point (0,0)

  • 0 > 0 - 1 = -1 correct
  • 0 < -1/2(0) + 2 = 2 - correct
  • Yes

Point (-4,1)

  • 1 > -4 - 1 = -5 - correct
  • 1 < -1/2(-4) + 2 = 4 - correct
  • Yes

Point (-1,2)

  • 2 > -1 - 1= -2 - correct
  • 2 < -1/2(-1) + 2 = 2.5 - correct
  • Yes

b) The system of inequalities

  • x - 2y ≥ 4
  • 2x - y ≥ -1

Point (0, 0)

  • 0 - 2(0) = 0 ≥ 4 - incorrect
  • 2(0) - 0 = 0  ≥ -1 - correct
  • No

Point (-4, 1)

  • -4 - 2(1) = - 6 ≥ 4 - incorrect
  • 2(-4) - 1 = - 9 ≥ -1  - correct
  • No

Point (-1,2)

  • -1 - 2(2) = -5  ≥ 4 - incorrect
  • 2(-1) - 2 = - 4 ≥ -1  - correct
  • No

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Answer:

  a.  all points are part of the solution set

  b.  no points are in the solution set

Step-by-step explanation:

Graphs of the two systems of inequalities are attached.

a) The red area on the graph is the solution. The two inequalities can be combined to one compound inequality:

  x -1 < y < -1/2x +2

The two boundary lines intersect at x-1 = -1/2x +2   ⇒   x = 2, so there are no points in the solution set with x-values of 2 or greater.

All of the given points are in the solution set.

__

b) The green area of the graph is the solution. Solving each inequality for y gives ...

  x -2y ≥ 4  ⇒  1/2x -2 ≥ y

  2x -y ≥ -1  ⇒  2x +1 ≥ y

The allowed values of y must be less than or equal to the minimum of these limits. The limits have the same value at 1/2x -2 = 2x +1   ⇒   x = -2, so the solution set is the area below and right of that intersection point of the two boundary lines.

None of the given points are in the solution set.

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