Consider the following system of linear inequalities:
\(x - y \le 3\) and \(x + y \ge 5\)
The solution of the inequalities lies in
both first quadrant and second quadrant only
The inequality \(x - y \le 3\) is satisfied on the side of its boundary containing the origin, while \(x + y \ge 5\) is satisfied on the side not containing the origin. In the third quadrant both \(x,y<0\), so \(x+y\) can never reach \(5\); in the fourth quadrant, \(x+y\ge5\) with \(y<0\) forces \(x-y\) to exceed \(3\). Testing points such as \((5,5)\) and \((-1,6)\) confirms the solution region lies only in the first and second quadrants.
Similar Questions
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?
Important Questions from Inequalities in two Variable
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?
