GRAPHING SYSTEMS OF INEQUALITIES

Graph the system of linear inequalities.

Example 1 :

x < 5

x > -4

Solution :

Given,

x < 5 and x > -4

x  =  5 and x  =  -4

By plotting the x values on the graph, we get

If x  =  -3, then -3 > -4 (True)

So, we can shade the region which is right to x > -4

If x  =  4, then 4 < 5 (True)

So, we can shade the region which is left to x < 5.

Example 2  :

y > -2

y ≤ 1

Solution  :

Given,

y > -2 and y ≤ 1

y  =  -2 and y  =  1

By plotting the y values on the graph, we get

If y  =  0, then 0 ≤ 1 (True)

So, we can shade the region which is below to y ≤ 1

If y  =  0, then 0 > -2 (True)

So, we can shade the region which is above to y > -2.

Example 3 :

x ≥ 0

x + y < 11

Solution :

Given,

x ≥ 0, and x + y < 11

x  =  0

x + y  =  11

To graph the line x + y  =  11, we find x and y intercepts.

The point on the line x + y  =  11 are (11, 0) and (0, 11)

By plotting the points on the graph, we get

If x  =  0, then 2 ≥ 0 (True)

So, we can shade the region which is right to x ≥ 1

If x  =  5 and y  =  2, then 5 + 2 < 11 (True)

So, we can shade the region which is below to x + y < 11.

Example 4 :

x + y ≥ -2

-5x + y < -3

Solution :

Given,

x + y ≥ -2 and -5x + y < -3

Converting the inequalities into equations.

x + y  =  -2 and -5x + y  =  -3

To graph the line x + y  =  -2, we find x and y intercepts.

To graph the line -5x + y  =  -3, we find x and y intercepts.

The point on the line x + y  =  -2 are (-2, 0) and (0, -2)

The point on the line -5x + y  =  -3 are (3/5, 0) and (0, -3)

By plotting the points on the graph, we get

If x  =  0 and y  =  -1, then 0 - 1 ≥ -2 (True)

So, we can shade the region which is above to x + y ≥ -2

If x  =  2 and y  =  -1, then -5(2) - 1 < -3 (True)

So, we can shade the region which is right to -5x + y < -3.

Example 5 :

y ≥ -4

y < -2x + 10

Solution :

Given,

y ≥ -4 and y < -2x + 10

Converting the inequalities into equations.

y  =  -4 and y  =  -2x + 10

To graph the line y  =  -2x + 10, we find x and y intercepts.

The point on the line y  =  -2x + 10 are (5, 0) and (0, 10)

By plotting the points on the graph, we get

If y  =  -1, then -1 ≥ -4 (True)

So, we can shade the region which is above to y ≥ -4

If x  =  2 and y  =  2, then 2 < -2(2) + 10 (True)

So, we can shade the region which is below to y < -2x + 10.

Example 6 :

y > 2x - 7

4x + 4y < -12

Solution :

Given,

y > 2x – 7 and 4x + 4y < -12

 y  =  2x - 7 and 4x + 4y  =  -12

To graph the line y  =  2x - 7, we find x and y intercepts.

To graph the line 4x + 4y  =  -12 , we find x and y intercepts.

The point on the line y  =  2x - 7 are (7/2, 0) and (0, -7)

The point on the line 4x + 4y  =  -12 are (-3, 0) and (0, -3)

By plotting the points on the graph, we get

If x  =  -4 and y  =  -4, then -4 > 2(-4) - 7 (True)

So, we can shade the region which is left to y > 2x – 7

If x  =  -4 and y  =  -4, then 2 < 4(-4) + 4(-4) < -12 (True)

So, we can shade the region which is below to 4x + 4y < -12

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