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Consider the inequations 5x - 4y + 12 < 0, x + y < 2, x < 0 and y > 0. Which one of the following points lies in the common region?

A.
(0, 0)
B.
(-2, 4)
C.
(-1, 4)
D.
(-1, 2)

Solution:

Calculation:Converting the inequations to equations, we obtain:5x - 4y + 12 = 0, x + y = 2, x = 0, y = 0.For 5x - 4y + 12 = 0Putting x = 0, 0 - 4y = - 12 ? y = 3Putting y = 0, 5x - 0 = - 12 ? x = - 2.4 This line meets the x-axis at (-2.4, 0) and the y-axis at (0, 3). Draw a line through these points.We see that the origin (0, 0) does not satisfy the inequation 5x - 4y + 12 < 0Therefore, the region that does not contain the origin is the solution of the inequality 5x - 4y + 12 < 0For x + y = 2Putting x = 0, y = 2Putting y = 0, x = 2 This line meets the x-axis at (2, 0) and the y-axis at (0, 2). Draw a line through these points.We see that the origin (0, 0) does satisfy the inequation x + y < 2Therefore, the region contains the origin is the solution of the inequality x + y < 2The solution to the inequalities is the intersection of the 5x - 4y + 12 < 0, x + y < 2, x < 0 and y > 0. Thus, the shaded region represents the solution set of the given set of inequalities.Now solving 5x - 4y + 12 = 0 and x + y = 2, we get the point P(-0.444, 2.444).Now, putting (-1, 2) in 5x - 4y + 12 < 0, - 5 - 8 + 12 < 0 ? -1 < 0 which is true.Again, putting (-1, 2) in x + y < 2, -1 + 2 < 2 ? -1 < 2 which is true.? Point (-1, 2) satisfies 5x - 4y + 12 = 0 and x + y = 2? (-1, 2) lies in the common region.Mistake Points Now, putting (0, 0) in 5x - 4y + 12 < 0, 0 - 0 + 12 < 0 ? 12 < 0 which is False.? Point (0, 0) not satisfies 5x - 4y + 12 = 0 and x + y = 2

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