Linear inequality of the given graph is given by option (A) and (C) .
- Graphing inequalities is the process of showing which parts of a number line contain values that "satisfy" a given inequality. This inequality plot shows the numbers you can use to replace x and y in the inequality to make the true statement.
(A) y > x /6 +2 ..(1) this line is represented by upper line .
and y ≤ x /4 + 1 ....(2) this line is represented by lower line .
Let consider a point (2, 3)
Put in equation (1) , we get
3 > 0.33 +2
3 > 2.33
which is true so shading will be in lower part of this line .
Put in equation (2) , we get
3 ≤ 0.5 +1
3 ≤ 1.5
which is false so shading will be in opposite direction in upper part of this line.
Hence , there is a common shading area between the lines which is required area .
(b) y < x /6 +2 ..(1) this line is represented by upper line and y ≥ x /4 + 1 ....(2) this line is represented by lower line . Let consider a point (2, 3)
Put in equation (1) , we get
3 < 0.33 +2
3 < 2.33
which is false so shading will be in opposite direction to the point in upper part of this line .
Put in equation (2) , we get
3 ≥ 0.5 +1
3 ≥ 1.5
which is true so shading will be in lower part of this line.
Hence , there is no common shading area between the lines .
(C) y ≥ x /6 +2 ..(1) this line is represented by upper line and y < x /4 + 1 ....(2) this line is represented by lower line . Let consider a point (2, 3)
Put in equation (1) , we get
3 ≥ 0.33 +2
3 ≥ 2.33
which is true so shading will be in direction to the point in lower part of this line .
Put in equation (2) , we get
3 < 0.5 +1
3 < 1.5
which is is false so shading will be in opposite direction to the point in upper part of this line
Hence , there is a common shading area between the lines which is required area .
(D) y ≤ x /6 +2 ..(1) this line is represented by upper line and y > x /4 + 1 ....(2) this line is represented by lower line . Let consider a point (2, 3)
Put in equation (1) , we get
3 ≤ 0.33 +2
3 ≤ 2.33
which is false so shading will be in opposite direction to the point in upper part of this line .
Put in equation (2) , we get
3 > 0.5 +1
3 > 1.5
which is true so shading will be in lower part of this line.
Hence , there is no common shading area between the lines .
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